Full Text Transcript
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(cid:1)(cid:2)(cid:3)(cid:2)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)
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LLLLLEEEEEAAAAARRRRRNNNNNIIIIINNNNNGGGGG OOOOOBBBBBJJJJJEEEEECCCCCTTTTTIIIIIVVVVVEEEEESSSSS
After studying this chapter, you will be able to:
(cid:1) Know the concept of limits and continuity;
(cid:1) Understand the theoruems underlying limits and their applications; and
(cid:1) Know how to solve the problems relating to limits and continuity with the help of given
illustrations.
88888.....11111 IIIIINNNNNTTTTTRRRRROOOOODDDDDUUUUUCCCCCTTTTTIIIIIOOOOONNNNN
Intuitively we call a quantity y a function of another quantity x if there is a rule (method
procedure) by which a unique value of y is associated with a corresponding value of x.
AAAAA fffffuuuuunnnnnccccctttttiiiiiooooonnnnn is defined to be rule that associates to any given number x a single number f(x)
to be read as function of x. f(x) does not mean f times x. It means given x, the rule f results
the number f(x).
Symbolically it may be written in the form y = f(x).
In any mathematical function y = f(x) we can assign values for x arbitrarily; consequently
x is the independent variable while the variable y is dependent upon the values of the
independent variable and hence dependent variable.
EEEEExxxxxaaaaammmmmpppppllllleeeee 11111::::: Given the function f(x) = 2x + 3 show that f(2x) = 2 f(x) – 3.
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: LHS. f(2x) = 2 (2x) + 3 = 4x + 6 – 3 = 2(2x + 3) – 3
= 2 f(x) – 3.
f(x+h)-f(x)
EEEEExxxxxaaaaammmmmpppppllllleeeee 22222::::: If f(x) = ax2 + b find .
h
f(x+h)-f(x) a(x+h)2+b-ax2-b a (x2+2xh+h2-x2) ha(2x+h)
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: = = =
h h h h
= a(2x + h)
NNNNNooooottttteeeee::::: f (x) = | x – a | means f (x) = x – a for x > a
= a – x for x < a.
= x – a for x = a
EEEEExxxxxaaaaammmmmpppppllllleeeee 33333::::: If f(x) = |x| + |x – 2| then redefine the function. Hence find f (3.5), f (– 2),
f(1.5).
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: If x > 2 f (x) = x + x – 2 = 2x – 2
If x < 0 f (x) = – x – x + 2 = 2 – 2x
If 0 < x < 2. f (x) = x – x + 2 = 2
So the given function can be redefined as
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f(x) = 2 – 2x for x < 0
= 2 for 0 ≤ x ≤ 2
= 2x – 2 for x > 2
for x = 3.5 f (x) = 2(3.5) – 2 = 5 , f (3.5) = 5
for x = – 2 f (x) = 2 – 2(– 2) = 6 f (–2) = 6
for x = 1.5 f (x) = 2. f (1.5) = 2
NNNNNooooottttteeeee..... Any function becomes undefined (i.e. mathematically cannot be evaluated) if
denominator is zero.
x+1
EEEEExxxxxaaaaammmmmpppppllllleeeee 44444::::: If f(x) = find f(0), f(1), f(– 1).
x2−3x−4
x+ 1 1 -1 2 1 0
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: f(x) = ∴f(0) = = , f(1) = = - f(–1) = which is
(x-4)(x+1) -4 4 (−3)(2) 3 0
not possible
i.e. it is undefined.
EEEEExxxxxaaaaammmmmpppppllllleeeee 55555::::: If f(x) = x2 – 5 evaluate f(3), f(–4), f(5) and f(1)
SSSSSooooollllluuuuutttttiiiiiooooonnnnn::::: f(x) = x2 – 5
f(3) = 32 – 5 = 9 – 5 = 4
f(–4) = (– 4)2 – 5 = 16 – 5 = 11
f(5) = 52 – 5 = 25 – 5 = 20
f(1) = 12 – 5 = 1 – 5 = – 4
88888.....22222 TTTTTYYYYYPPPPPEEEEESSSSS OOOOOFFFFF FFFFFUUUUUNNNNNCCCCCTTTTTIIIIIOOOOONNNNNSSSSS
EEEEEvvvvveeeeennnnn aaaaannnnnddddd ooooodddddddddd fffffuuuuunnnnnccccctttttiiiiiooooonnnnnsssss ::::: if a function f(x) is such that f(–x) = f(x) then it is said to be an
even function of x.
EEEEExxxxxaaaaammmmmpppppllllleeeeesssss ::::: f (x) = x2 + 2x4
f (–x) = (–x ) 2 + 2 (–x )4 = x2 + 2x4 = f(x)
Hence f(x) = x2 + 2x4 is an even function.
On the other hand if f(x) = – f(x) then f(x) is said to be an odd function.
EEEEExxxxxaaaaammmmmpppppllllleeeeesssss ::::: f (x) = 5x + 6x3
f (-x) = 5(-x ) + 6(-x )3 = -5x - 6x3 = -(5x + 6x3)
Hence 5x + 6x3 is an odd function.
PPPPPeeeeerrrrriiiiiooooodddddiiiiiccccc fffffuuuuunnnnnccccctttttiiiiiooooonnnnnsssss::::: A function f (x) in which the range of the independent variable can be
separated into equal sub intervals such that the graph of the function is the same in each
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part then it is periodic function. Symbolically if f(x + p) = f(x) for all x, then p is the period
of f.
IIIIInnnnnvvvvveeeeerrrrrssssseeeee fffffuuuuunnnnnccccctttttiiiiiooooonnnnn::::: If y = f(x) defined in an interval (a, b) is a function such that we express
x as a function of y say x = g(y) then g(y) is called the inverse of f(x)
5x+3 3-9y
Example: i) if y= , then x= is the inverse of the first function.
2x+9 2y-5
ii) x=3 y is the inverse function of y=x3.
CCCCCooooommmmmpppppooooosssssiiiiittttteeeee FFFFFuuuuunnnnnccccctttttiiiiiooooonnnnn::::: If y = f(x) and x = g(u) then y = f {g(u)} is called the function of a
function or a composite function.
⎛ ⎞
Example : If a function f(x) = log 1 1 + -x x prove that f(x 1 ) + f(x 2 ) = f ⎜ ⎜ ⎜⎜⎝1 x + 1 x +x x 2 ⎠ ⎟ ⎟ ⎟ ⎟
1 2
1+x 1+x
1 2
Solution : f(x )+f(x ) = log + log
1 2 1-x 1-x
1 2
1+x 1+x
1× 2
= log
1-x 1-x
1 2
x +x
1+ 1 2 ⎛ ⎞
1+ x + x + x x 1+x x ⎜ x +x ⎟
= log 1 –x 1 –x 2 + x 1 x 2 =log x + 1 x 2 =f⎜ ⎜⎜⎝1+ 1 x x 2 ⎠ ⎟ ⎟ ⎟ . Proved
1 2 1 2 1- 1 2 1 2
1+x x
1 2
EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 88888(((((AAAAA)))))
CCCCChhhhhoooooooooossssseeeee ttttthhhhheeeee mmmmmooooosssssttttt aaaaapppppppppprrrrroooooppppprrrrriiiiiaaaaattttteeeee oooooppppptttttiiiiiooooonnnnn (((((aaaaa))))) (((((bbbbb))))) (((((ccccc))))) ooooorrrrr (((((ddddd)))))
1. Given the function f(x) = x2 – 5, f( 5)is equal to
a) 0 b) 5 c) 10 d) none of these
5x+1
2. If f(x)= then f(x) is
5x-1
a) an even function b) an odd function
c) a composite function d) none of these
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3. If g(x) = 3 – x2 then g(x) is
a) an odd function b) a periodic function
c) an even function d) none of these
q×(x-p) p×(x-q)
f(x)= +
4. If then f(p) + f(q) is equal to
(q-p) (p-q)
a) p +q b) f(pq) c) f(p – q) d) none of these
5. If f(x) = 2x2 – 5x + 4 then 2f(x) = f(2x) for
a) x=1 b) x = – 1 c) x = ± 1 d) none of these
6. If f(x) = logx (x > 0) then f(p) + f(q) +f(r) is
a) f(pqr) b) f(p)f(q)f(r) c) f(1/pqr) d) none of these
+ −
f(4 h) f(4)
7. If f(x) = 2x2 – 5x +2 then the value of is
h
a) 11 – 2h b) 11 + 2h c) 2h – 11 d) none of these
px-q
8. If y=h(x)= then x is equal to
qx-p
a) h(1/y) b) h (–y) c) h(y) d) none of these
9. If f(x) = x2 – x then f( h+1) is equal to
a) f(h) b) f(–h) c) f(–h + 1) d) none of these
1-x
10. If f(x)= then f (f(1/x)) is equal to
1+x
a) 1/x b) x c) –1/x d) none of these
88888.....33333 CCCCCOOOOONNNNNCCCCCEEEEEPPPPPTTTTT OOOOOFFFFF LLLLLIIIIIMMMMMIIIIITTTTT
IIIII))))) We consider a function f(x) = 2x. If x is a number approaching to the number 2 then f(x) is
a number approaching to the value 2 × 2 = 4.
The following table shows f(x) for different values of x approaching 2
x f(x)
1.90 3.8
1.99 3.98
1.999 3.998
1.9999 3.9998
2 4
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Here x approaches 2 from values of x<2 and for x being very close to 2 f(x) is very close to
4. This situation is defined as left-hand limit of f(x) as x approaches 2 and is written as lim
→
f(x) = 4 as x 2 –
Next
x f(x)
2.0001 4.0002
2.001 4.002
2.01 4.02
2.0 4
Here x approaches 2 from values of x greater than 2 and for x being very close to 2 f(x) is
very close to 4. This situation is defined as right–hand limit of f(x) as x approaches 2 and
is written as lim f(x) = 4 as x→2 +
So we write
lim f(x) = lim f(x) = 4
x→2− x→2+
Thus lim f(x) is said to exist when both left-hand and right-hand limits exists and they
x→a
are equal. We write as
lim f(x)= lim f(x)= lim f(x)
x→a− x→a+ x→a
Thus, if lim f (a+h) = lim f (a–h) , (h>o)
(cid:2) (cid:2)
h o h o
then lim f(x) exists
(cid:2)
x a
We now consider a function defined by
⎧⎪
⎪2x-2 for x<0
⎪
f(x)=⎨ 1 for x=0
⎪
⎪
⎪⎩2x+2 for x>0
We calculate limit of f(x) as x tend to zero. At x = 0 f(x) = 1 (given). If x tends to zero from
left-hand side for the value of x<0 f(x) is approaching (2×0) –2 = –2 which is defined as
→
left-hand limit of f(x) as x 0 - we can write it as
Thus lim =−2
x→0−
Similarly if x approaches zero from right-hand side for values of x>0 f(x) is approaching 2
× 0 + 2 = 2. We can write this as lim f(x) = 2.
x→0+
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In this case both left-hand limit and right-hand exist but they are not equal. So we may
conclude that limf(x) does not exist.
x® 0
88888.....44444 UUUUUSSSSSEEEEEFFFFFUUUUULLLLL RRRRRUUUUULLLLLEEEEESSSSS OOOOOFFFFF TTTTTHHHHHEEEEEOOOOORRRRREEEEEMMMMMSSSSS OOOOONNNNN LLLLLIIIIIMMMMMIIIIITTTTTSSSSS
Let limf(x) = (cid:2) and limg(x) = m
x→a x→a
(cid:2)
where and m are finite quantities
(cid:2)
i) lim{f(x) + g(x)} = limf(x) + lim g(x) = + m
x→a x→a x→a
That is limit of the sum of two functions is equal to the sum of their limits.
ii) lim{f(x) – g(x)} = limf(x) – limg(x) = (cid:2) –m
x→a x→a x→a
That is limit of the difference of two functions is equal to difference of their limits.
iii) lim{f(x) . g(x)} = limf(x) . limg(x) = (cid:2) m
x→a x→a x→a
That is limit of the product of two functions is equal to the product of their limits.
iv) lim{f(x)/g(x)} = {limf(x)}/{limg(x)} = (cid:2) /m
x→a x→a x→a
That is limit of the quotient of two functions is equal to the quotient of their limits.
v) limc = c where c is a constant
x→a
That is limit of a constant is the constant.
vi) limcf(x) = climf(x)
x→a x→a
= =
limF{f(x)} F{lim f(x)} F(l)
vii)
x→a x→a
viii) lim 1 = lim 1 → +∞ (h>0)
– –
x h
x→
0 + h
→
0
lim 1 = lim 1 → –∞ (h>0)
– –
x -h
x→ 0– h → 0
∞ is a very-very large number called infinity
Thus lim 1–x does not exist.
x→
0
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:24)
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1 1
Example 1: Evaluate: (i)lim(3x + 9); (ii) lim (iii) lim
x→2 x→5 x − 1 x→a x − a
+ = + = + =
Solution: (i) lim(3x 9) 3.2 9 (6 9) 15
x→2
1 1 1
= =
(ii) lim
x→5 x − 1 5 − 1 4
1 1 → + ∞ 1 → ∞
(iii) l x i → m a x − a does not exist, x→ lim a+ x-a and x l → im a - x-a -
⎛ ⎞
1
[Hint: L.H.S. = h li → m 0 and h li → m 0 ⎜ ⎝-h ⎟ ⎠ (h>o)
x
2−5x+6
Example 2: Evaluate lim .
x→2 x−2
Solution: At x = 2 the function becomes undefined as 2-2 = 0 and division by zero is not
mathematically defined.
So lim { x2− 5x + 6/(x − 2) } = lim { (x − 2)(x − 3)/(x − 2) }= lim(x − 3) (∵ x-2 ≠ 0)
x→2 x→2 x→2
= 2-3 = -1
2+ −
x 2x 1
Example 3: Evaluate lim .
→
x 2 2+
x 2
2+ − 2+ −
x 2+ 2x − 1 = x l → im 2 (x 2x 1) = x l → im 2 x x l → im 2 2x 1
Solution: lim
→
x 2 2+ 2+ 2+
x 2 lim x 2 lim x 2
→ →
x 2 x 2
2+ × −
(2) 2 2 1 7
= =
2+ 6
(2) 2
88888.....55555 SSSSSOOOOOMMMMMEEEEE IIIIIMMMMMPPPPPOOOOORRRRRTTTTTAAAAANNNNNTTTTT LLLLLIIIIIMMMMMIIIIITTTTTSSSSS
We now state some important limits
(ex-1)
a) lim =1
x→0 x
(cid:10)(cid:11)(cid:10) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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ax-1
b) lim =log a (a>0)
x→0 x e
log(1+x)
c) lim =1
x→0 x
d) lim ⎛ ⎜ ⎜⎜⎝ 1+ 1⎟ ⎟ ⎟ ⎞ ⎠ x =e or lim (1+x)x 1 =e
x→∝ x x→0 x
xn-an
e) lim =nan-1
x→a x-a
(1+x)n-1
f) lim =n
x→0 x
(A) The number e called exponential number is given by e = 2.718281828 —— = 2.7183. This
number e is one of the useful constants in mathematics.
(B) In calculus all logarithms are taken with respect to base ‘e’ that is log x=log x.
e
ILLUSTRATIVE EXAMPLES
x2-6x+9 x2 −6x+9
Example 1: Evaluate: lim , where f(x) = . Also find f (3)
x→3 x-3 x−3
Solution: At x = 3 the function is undefined as division by zero is meaningless. While taking
→ →
the limit as x 3 the function is defined near the number 3 because when x 3 x
≠
cannot be exactly equal to 3 i.e. x – 3 0 and consequently division by x – 3 is
permissible.
x2-6x+9 (x-3)2 0
= =
Now lim lim lim (x-3) =3-3 =0. f(3) = is undefined
x→3 x-3 x→3 x-3 x→3 0
The reader may compute the left-hand and the right-hand limits as an exercise.
Example 2: A function is defined as follows:
⎧⎪
⎪-3x whenx<0
f(x)=⎨
⎪
⎪⎩2x whenx>0
lim f(x).
Test the existence of
x→0
Solution: For x approaching 0 from the left x < 0.
Left-hand limit = lim f(x) = lim (– 3x) = 0
x→0- x→0-
When x approaches 0 from the right x > 0
Right-hand limit = lim f(x) = lim 2x = 0
x→0+ x→0+
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:25)
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(cid:1)(cid:2)(cid:3)(cid:2)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:6)(cid:10)(cid:11)(cid:8)(cid:4)(cid:2)(cid:8)(cid:12)(cid:2)(cid:4)(cid:13)(cid:14)(cid:2)(cid:8)(cid:4)(cid:12)(cid:2)(cid:4)(cid:2)(cid:15)(cid:16)(cid:6)(cid:7)(cid:17)(cid:17)(cid:18)(cid:11)(cid:7)(cid:10)(cid:19)
Since L.H. limit = R.H. Limit, the limit exists. Thus, lim f(x) = 0.
x→0
1
Example 3: Does lim exist ?
x→π Π-x
1 1
→ ∞ ∞
Solution: lim = and lim =+ ;
x→π+0 π -x x→π-0 π-x
⎛ 1 ⎞ ⎡ 1 ⎤ ⎛ 1 ⎞
R.H.L. x l → im π ⎜ ⎝Π−x ⎟ ⎠ = h li → m 0 ⎢ ⎣Π−(Π+h ⎥ ⎦ = h li → m 0 ⎜ ⎝ − h ⎟ ⎠ →−∞
Since the limits are unequal the limit does not exist.
⎛ 1 ⎞ ⎡ 1 ⎤ ⎛1⎞
R.H.L. = x l → im π ⎜ ⎝Π−x ⎟ ⎠ = h li → m v ⎢ ⎣Π−(Π−h ⎥ ⎦ = h li → m 0 ⎜ ⎝h ⎟ ⎠ →+∞
2
x +4x+3
Example 4: : lim .
x→3
x
2
+6x+9
2 2
x +4x+3 x +3x+x+3 x(x+3)+1(x+3) (x+3)(x+1) x+1
Solution: = = = =
2 2 2 2
x +6x+9 (x+3) (x+3) (x+3) x+3
2
x +4x+3 x+1 4 2
∴ lim = lim = = .
x → 3 x 2 +6x+9 x → 3x+3 6 3
Example 5: Find the following limits:
x-3 x+h- x
(i) x li → m 9 x-9 ; (ii) h l → im 0 h if h > 0.
Solution:
x-3 x-3 1 x-3 1 1
(i) = = . ∴ lim = li → m = .
x-9 ( x+3) ( x-3) x+3 x → 9 x-9 x 9 x+3 6
x+h- x x+h-x 1 1
(ii) = = ∴ lim x+h- x = lim
h h ( x+h+ x) x+h+ x h → 0 h h→0 x+h+ x
1 1 1
= =
= lim x+h+ lim x x+ x 2 x .
h→0 h→0
3x+x
lim
Example 6: Find → .
x 0 7x-5x
3x+|x| 3x+x
Solution: Right-hand limit = lim = lim = lim 2 = 2
→ → →
x 0+7x-5|x| x 0+7x-5x x 0+
(cid:10)(cid:11)(cid:26)(cid:27) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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3x+ x 3x-(x) 1 1
Left-hand limit lim = lim = lim = .
x → 0- 7x-5 x x → 0- 7x-5 (-x) x → 0- 6 6
≠
Since Right-hand limit Left-hand limit the limit does not exist.
x -x
e -e
Example 7: Evaluate lim
→
x 0 x
x -x x -x x -x
e -e (e -1)-(e -1) e -1 e -1
Solution: lim = lim = lim - lim =1-1=0
→ → → →
x 0 x x 0 x x 0 x x 0 x
⎛ ⎞x
⎜ 9⎟
Example 8: Find
x
→ lim ∝ ⎜⎜⎝ 1+
x
⎟ ⎟⎠ . (Form 1`)
x x
Solution: It may be noted that 9 approaches ∝ as x approaches ∞. i.e. x l → im ∞ 9 →∞
⎧⎪
⎪⎛ ⎞x/9
⎫⎪
⎪
9
⎪⎜ ⎟ ⎪
x l → im ∞ ⎛ ⎜ ⎜⎜⎝ 1+ x 9⎟ ⎟ ⎟ ⎞ ⎠ x = x/ l 9 im →∝ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪
⎪
⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝ 1+ x 1 ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪
⎪
⎪⎩ 9 ⎪⎭
⎧⎪⎛ ⎞z⎫⎪ 9
Substituting x/9 = z the above expression takes the form z l → im ∝ ⎪ ⎨ ⎪ ⎪⎩ ⎜ ⎜⎜⎝ 1+ z 1 ⎟⎠ ⎟ ⎟ ⎪ ⎬ ⎪⎭ ⎪
⎧⎪ ⎛ ⎞z⎫⎪ 9
= ⎪ ⎨ ⎪ ⎪⎩z→ lim ∞ ⎜ ⎜⎜⎝ 1+
z
1⎟ ⎟ ⎟⎠ ⎪ ⎬
⎪⎭
⎪ =e 9 .
2x+1 ⎡ ∞⎤
Example 9: Evaluate: x l → im ∝ 3 . ⎢ ⎣ Form ∞ ⎥ ⎦
x +1
Solution: As x approaches ∝ 2x + 1 and x3 + 1 both approach ∝ and therefore the given
∝
function takes the form which is indeterminate. Therefore instead of evaluating directly let
∝
us try for suitable algebraic transformation so that the indeterminate form is avoided.
⎛ ⎞
2 1 ⎜ 2 1 ⎟ 2 1
x 2 + x 3 x l → im ∝ ⎜ ⎜⎝ x 2 + x 3 ⎟ ⎟⎟⎠ x l → im ∝ x 2 + x l → im ∝ x 3 0+0 0 =
lim = = = = 0.
→∝ ⎛ ⎞
x 1 ⎜ 1 ⎟ 1 1+0 1
1+ x 3 x l → im ∝ ⎜ ⎜⎝ 1+ x 3⎠ ⎟ ⎟⎟ x l → im ∝ 1+ x l → im ∝ x 3
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:26)(cid:26)
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(cid:1)(cid:2)(cid:3)(cid:2)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:6)(cid:10)(cid:11)(cid:8)(cid:4)(cid:2)(cid:8)(cid:12)(cid:2)(cid:4)(cid:13)(cid:14)(cid:2)(cid:8)(cid:4)(cid:12)(cid:2)(cid:4)(cid:2)(cid:15)(cid:16)(cid:6)(cid:7)(cid:17)(cid:17)(cid:18)(cid:11)(cid:7)(cid:10)(cid:19)
2 2 2 2
1 +2 +3 +..........+x
Example 10: Findlim
x→∞
x
3
2 2 2 2
1 +2 +3 +..........+x
Solution: lim
x→∝ 3
x
[x(x+1)(2x+1)] 1
⎧⎪⎪ ⎛
⎜ 1
⎞
⎟
⎛
⎜ 1
⎞
⎟
⎫⎪⎪
x l → im ∝ 6x 3 = 6x l → im ∝ ⎨ ⎪ ⎪⎩ ⎜⎜⎝ 1+ x ⎟ ⎟⎠ ⎜⎜⎝ 2+ x ⎟ ⎟⎠⎪ ⎬ ⎪⎭
1 1
= ×1×2= .
6 3
⎛ ⎞
⎜ 1 2 3 n ⎟
Example 11:
x
li
→
m
∝
⎜ ⎜⎝
1-n 2
+
1-n 2
+
1-n 2
.......................+
1-n 2
⎟ ⎟⎟⎠
⎛ ⎞
⎜ 1 2 3 n ⎟
Solution : =
x
li
→
m
∝
⎜ ⎜⎝
1-n 2
+
1-n 2
+
1-n 2
.......................+
1-n 2
⎟ ⎟⎟⎠
1
= lim (1+2+3 ………..+n)
x→∝ 1-n2
1 n(n+1)
lim
= ×
x→∝ 1-n2 2
1 n(n+1)
lim
= ×
x→∝ 1-n2 2
1 n
lim
=
2 x→∝ 1-n
⎛ ⎞
⎜ ⎟
1 ⎜ ⎜ ⎜ 1 ⎟ ⎟ ⎟ ⎟
= lim ⎜ ⎜ 1 ⎟ ⎟
2 x→∝ ⎜ ⎜⎝ -1⎠ ⎟ ⎟
n
⎛ ⎞
1 1 1 ⎜ 1⎟
= lim = (–1) = ⎜⎜⎝ - ⎟ ⎟⎠
2 x→∝ 0-1 2 2
Exercise 8 (B)
Choose the most appropriate option (a) (b) (c) or (d)
1. lim f (x) when f(x) = 6 is
x→0
a) 6 b) 0 c) 1/6 d) none of these
(cid:10)(cid:11)(cid:26)(cid:12) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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2. lim (3x + 2) is equal to
x→2
a) 6 b) 4 c) 8 d) none of these
2
x -4
3. lim is equal to
x→-2 x+2
a) 4 b) –4 c) does not exist d) none of these
⎛ ⎞
⎜ 3 ⎟
4. l
x
i
→
m
∝
⎜ ⎜⎝
x 2
+2 ⎟ ⎟⎟⎠
a) 0 b) 5 c) 2 d) none of these
5. lim log e x is evaluated to be
x→1
a) 0 b) e c) 1 d) none of these
6. The value of the limit of f(x) as x → 3 when f(x) =ex2+2x+1 is
a) e 15 b) e16 c) e10 d) none of these
⎛ ⎞
7.
x
l
→
im
1/2
⎜ ⎜ ⎜ ⎜⎝6x 8 2 x
-
3
5
-
x
1
+1
⎟ ⎟ ⎟ ⎟⎟⎠ is equal to
a) 5 b) –6 c) 6 d) none of these
2 2
1+2x - 1-2x
8. lim is equal to
x→0 x 2
a) 2 b) –2 c) ½ d) none of these
x-q- p-q
9. lim (p>q) is evaluated as
x→p
x
2
-p
2
1 1 1
a) b) c) d) none of these
p p-q 4p p-q 2p p-q
x
(3 -1)
10. lim is equal to
x→0 x
a) 10 3 log 3 b) log e c) log 3 d) none of these
10 3 e
x x
5 +3 -2
11. lim will be equal to
x→0 x
a) log 15 b) log (1/15) c) log e d) none of these
e
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:26)(cid:20)
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(cid:1)(cid:2)(cid:3)(cid:2)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:6)(cid:10)(cid:11)(cid:8)(cid:4)(cid:2)(cid:8)(cid:12)(cid:2)(cid:4)(cid:13)(cid:14)(cid:2)(cid:8)(cid:4)(cid:12)(cid:2)(cid:4)(cid:2)(cid:15)(cid:16)(cid:6)(cid:7)(cid:17)(cid:17)(cid:18)(cid:11)(cid:7)(cid:10)(cid:19)
10x − 5x-2x
12. lim is equal to
x→0 x2
a) log 2 + log 5 b) log 2 log 5 c) log 10 d) none of these
e e e e e
f(x+h)-f(x)
13. If f(x) = ax2 + bx+c then lim is equal to
x→0 h
a) ax +b b) ax + 2b c) 2ax +b d) none of these
2x2-7x+6
14. lim is equal to
x→2 5x2-11x+2
a) 1/9 b) 9 c) –1/9 d) none of these
x3-5x2+2x+2
15. lim is equal to
x→1 x3+2x2-6x+3
a) 5 b) –5 c) 1/5 d) none of these
x3-t3
16. lim is evaluated to be
x→t x2-t2
⎛ ⎞
⎜3⎟
a) 3/2 b) 2/3t c) ⎜⎜⎝ ⎟ ⎟⎠t d) none of these
2
4x4+5x37x2+6x
17. lim is equal to
x→0 5x5+7x2+x
a) 7 b) 5 c) –6 d) none of these
(x2− 5x + 6 ) (x2 -3x +2)
18. lim is equal to
x→2 x3-3x2+4
a) 1/3 b) 3 c) –1/3 d) none of these
3x4 + 5x2 + 7x + 5
19. lim is evaluated
x→∝ 4x2
3
a) b) 3 c) –1/4 d) none of these
4
(ex + e -x - 2 ) (x2 -3x +2)
20. lim is equal to
x→0 (x-1)
a) 1 b) 0 c) –1 d) none of these
(cid:10)(cid:11)(cid:26)(cid:21) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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(1-x-1/3)
21. lim is equal to
x→1
(1-x-2/3)
a) –1/2 b) 1/2 c) 2 d) none of these
(x2-16)
lim
22. is evaluated as
x→4 (x-4)
a) 8 b) –8 c) 0 d) none of these
x2- x
23. lim is equal to
x→1 x-1
a) –3 b) 1/3 c) 3 d) none of these
x3 −1
24. lim is equal to
x→1 x−1
a) 3 b) –1/3 c) –3 d) none of these
(1+x)6
25. then lim f(x) is equal to
(1+x)2 − 1 x→0
a) –1 b) 3 c) 0 d) none of these
(1+px)
26. lim log is equal to
x→0 e3x-1
a) p/3 b) p c) 1/3 d) none of these
⎛ ⎞
1
27. lim ⎜ ⎟ is equal to
x→∞ ⎝ x3 +x2 +x+1⎠
a) 0 b) e c) –e6 d) none of these
2x2+7x+5
28. lim is equal to l where l is
x→∞ 4x2+3x-1
a) –1/2 b) 1/2 c) 2 d) none of these
(x x-m m)
29. lim is equal to
x→∞ 1-x-2/3
a) 1 b) –1 c) 1/ 2 d) none of these
(x+2)5/3-(p+2)5/3
30. lim is equal to
x→0 x-p
a) p b) 1/p c) 0 d) none of these
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:26)(cid:22)
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(cid:1)(cid:2)(cid:3)(cid:2)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:6)(cid:10)(cid:11)(cid:8)(cid:4)(cid:2)(cid:8)(cid:12)(cid:2)(cid:4)(cid:13)(cid:14)(cid:2)(cid:8)(cid:4)(cid:12)(cid:2)(cid:4)(cid:2)(cid:15)(cid:16)(cid:6)(cid:7)(cid:17)(cid:17)(cid:18)(cid:11)(cid:7)(cid:10)(cid:19)
x3+3x2-9x-2
31. If f(x) and limf(x) exists then lim(x) is equal to
x3-x-6 x→2 x→2
a) 15/11 b) 5/11 c) 11/15 d) none of these
5+2x-(3+2)
32. lim= is equal to
x→6 x2-6
3-2 1
a) 3 – 2 b) c) d) none of these
2-6 2-6
4-x2
33. lim is equal to
x→2 3- x2+5
a) 6 b) 1/6 c) –6 d) none of these
x3/2-23/4
lim
34. exists and is equal to a finite value which is
x→ 2 x-21/4
√
a) –5 b) 1/6 c) 3 2 d) none of these
⎛1⎞
35. lim ⎜ ⎟ log (1–x/2 is equal to
x→0 ⎝ x⎠
a) –1/2 b) 1/2 c) 2 d) none of these
(x− 1)2
lim
36. is equal to
x→1 (x− 1)(x2 − 1)
a) 1 b) 0 c) –1 d) none of these
⎡ 13+23+33+--+x3⎤
lim ⎢ ⎥
37. is equal to
x→∞ ⎣ x4 ⎦
a) 1/4 b) 1/2 c) –1/4 d) none of these
8.6 CONTINUITY
By the term “continuous” we mean something which goes on without interruption and
without abrupt changes. Here in mathematics the term “continuous” carries the same
meaning. Thus we define continuity of a function in the following way.
A function f(x) is said to be continuous at x = a if and only if
(i) f(x) is defined at x = a
lim
(ii) lim f(x) = f(x)
x→a-
x→a+
(cid:10)(cid:11)(cid:26)(cid:23) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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(iii) lim f(x) = f(a)
x→a
In the second condition both left-hand and right-hand limits exists and are equal.
In the third condition limiting value of the function must be equal to its functional value at
x = a.
Useful Information:
(i) The sum difference and product of two continuous functions is a continuous function.
This property holds good for any finite number of functions.
(ii) The quotient of two continuous functions is a continuous function provided the
denominator is not equal to zero.
1
Example 1 : f(x) = -x when 0< x < 1/2
2
3
= -x when ½ < x < 1
2
1 1
= when x =
2 2
Discuss the continuity of f(x) at x = ½.
Solution : lim f(x) = lim (1/2 –x) = 1/2 – 1/2 = 0
x→1
-
x→1
-
2 2
lim f(x) = lim (3/2 –x) = (3/2 – 1/2) = 1
x→1
+
x→1
+
2 2
≠ lim
Since LHL RHL f(x) does not exist
x→1/2
Moreover f(1/2 ) = 1/2
1
=
Hence f(x) is not continuous of x = 1/2 , i.e. f (x) is discontinuous at x
2
x2+2x+5
Example 2 : Find the points of discontinuity of the function f(x) =
x2-3x+2
x2+2x+5 x2+2x+5
=
Solution : f(x) =
x2-3x+2 (x-1)(x-2)
For x = 1 and x = 2 the denominator becomes zero and the function f(x) is undefined at
x = 1 and x = 2. Hence the points of discontinuity are at x = 1 and x = 2.
Example 3 : A function g(x) is defined as follows:
g(x) = x when 0< x < 1
≥
= 2 – x when x 1
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:26)(cid:24)
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(cid:1)(cid:2)(cid:3)(cid:2)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:6)(cid:10)(cid:11)(cid:8)(cid:4)(cid:2)(cid:8)(cid:12)(cid:2)(cid:4)(cid:13)(cid:14)(cid:2)(cid:8)(cid:4)(cid:12)(cid:2)(cid:4)(cid:2)(cid:15)(cid:16)(cid:6)(cid:7)(cid:17)(cid:17)(cid:18)(cid:11)(cid:7)(cid:10)(cid:19)
Is g(x) is continuous at x = 1?
Solution :
lim g(x) = limx = 1
x→1- x→1-
lim lim
g(x) = ( 2 –x) = 2 – 1 = 1
x→1+ x→1+
∴ lim g(x) = lim g (x) = 1
x→1-
x→1+
Moreover g(1) = 2 –1 = 1
So lim g(x) = g(1) = 1
x→1
Hence f(x) is continuous at x = 1.
Example 4: The function f(x) = (x2 – 9) / (x – 3) is undefined at x = 3. What value must be
assigned to f(3) if f (x) is to be continuous at x = 3?
Solution : When x approaches 3 x ≠ 3 i.e. x – 3 ≠ 0
(x-3)(x+3)
So lim f(x) = lim
x→3 x→3 (x-3)
= lim (x + 3) = 3 + 3 = 6
x→3
Therefore if f(x) is to be continuous at x = 3, f(3) = limf(x) = 6.
x→3
Example 5: Is the function f(x) = | x | continuous at x = 0?
Solution: We know | x | = x when x > 0
= 0 when x = 0
= –x when x < 0
Now lim f(x) = lim (–x) = 0 and lim f(x) = lim x = 0
x→0- x→0- x→0+ x→0+
Hence lim f(x) = 0 = f(0)
x→0
So f(x) is continuous at x = 0.
Exercise 8(C)
Choose the most appropriate option (a) (b) (c) or (d)
1. If f(x) is an odd function then
f(-x)+f(x)
a) is an even function
2
b) [| x | + 1 ] is even when [x] = the greater integer x <
(cid:10)(cid:11)(cid:26)(cid:10) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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f(x)+f(-x)
c) is neither even or odd
2
d) none of these.
2. If f(x) and g(x) are two functions of x such that f(x) + g(x) = ex and f(x) – g(x) = e –x then
a) f(x) is an odd function b) g(x) is an odd function
c) f(x) is an even function d) g(x) is an even function
2x2+6x-5
3. If f(x) = is to be discontinuous then
12x2+x-20
a) x = 5/4 b) x = 4/5 c) x = –4/3 d) none of these.
4. A function f(x) is defined as follows
f(x) = x2 when 0 < x <1
= x when 1 < x < 2
= (1/4) x3 when 2 < x < 3
Now f(x) is continuous at
a) x = 1 b) x = 3 c) x = 0 d) none of these.
3x+|x|
5. lim
7x-5|x|
x→0
a) exists b) does not exist c) 1/6 d) none of these.
(x+1)
6. If f(x) = then lim f(x) and f(-1)
6x2+3+3x
x® -1
a) both exists b) one exists and other does not exist
c) both do not exists d) none of these.
x2-1
7. lim is evaluated to be
3x+1- 5x-1
x® 1
a) 4 b) 1/4 c) –4 d) none of these.
8. lim ( x+h- x) / h where h→0 is equal to
1
a) 1/ 2 x b) 1/2x c) x /2 d)
2 x
9. Let f(x) = x when x >0
= 0 when x = 0
= – x when x < 0
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:26)(cid:25)
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(cid:1)(cid:2)(cid:3)(cid:2)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:6)(cid:10)(cid:11)(cid:8)(cid:4)(cid:2)(cid:8)(cid:12)(cid:2)(cid:4)(cid:13)(cid:14)(cid:2)(cid:8)(cid:4)(cid:12)(cid:2)(cid:4)(cid:2)(cid:15)(cid:16)(cid:6)(cid:7)(cid:17)(cid:17)(cid:18)(cid:11)(cid:7)(cid:10)(cid:19)
Now f(x) is
a) discontinuous at x = 0 b) continuous at x = 0
c) undefined at x = 0 d) none of these.
10. If f(x) = 5+3x for x > 0 and f(x) = 5 – 3x for x < 0 then f(x) is
a) continuous at x = 0
b) discontinuous and defined at x = 0
c) discontinuous and undefined at x = 0
d) none of these.
⎧⎪
⎪(x-1)2
⎫⎪
⎪
lim ⎨ +(x2-1)⎬
11. ⎪ ⎪
x→1 ⎪⎩ x-1 ⎪⎭
a) does not exist b) exists and is equal to two
c) is equal to 1 d) none of these.
4x+1-4
lim
12.
x→0 2x
a) does not exist b) exists and is equal to 4
c) exists and is equal to 4 log 2 d) none of these.
e
(x2-16)
13. Let f(x) = for x ≠ 4
(x-4)
= 10 for x = 4
Then the given function is not continuous for
(a) limit f(x) does not exist
(b) limiting value of f(x) for x→ 4 is not equal to its function value f(4)
(c) f(x) is not defined at x = 4
(d) none of these.
14. A function f(x) is defined by f(x) = (x–2)+1 over all real values of x, now f(x) is
(a) continuous at x = 2 (b) discontinuous at x = 2
(c) undefined at x = 2 (d) none of these.
15. A function f(x) defined as follows f(x) = x+1 when x < 1
= 3 – px when x > 1
The value of p for which f(x) is continuous at x = 1 is
(a) –1 (b) 1 (c) 0 (d) none of these.
16. A function f(x) is defined as follows :
(cid:10)(cid:11)(cid:12)(cid:27) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
f(x)= x when x < 1
= 1+x when x > 1
= 3/2 when x = 1
Then f(x) is
(a) continuous at x = ½ (b) continuous at x = 1
(c) undefined at x = ½ (d) none of these.
17. Let f(x) = x/|x|. Now f(x) is
(a) continuous at x = 0 (b) discontinuous at x = 0
(c) defined at x = 0 (d) none of these.
18. f(x)= x–1 when x > 0
= – ½ when x = 0
= x + 1 when x < 0
f(x) is
(a) continuous at x = 0 (b) undefined at x = 0
(c) discontinuous at x = 0 (d) none of these.
⎜
⎛
x+6⎟
⎞x+4
19. lim⎜⎜⎝ ⎟ ⎟⎠ is equal to
x→0 x+1
(a) 64 (b) 1/e5 (c) –e5 (d) none of these.
(e2x-1)
20. lim is equal to
x→0 x
(a) ½ (b) 2 (c) 0 (d) none of these.
ex+1
21. lim is evaluated to be
x→∞ ex+2
(a) 0 (b) –1 (c) 1 (d) none of these.
⎛ xn-3n ⎞
⎜ ⎟
22. If lim = 108 then the value of n is
x→3 ⎝ x-3 ⎠
(a) 4 (b) –4 (c) 1 (d) none of these.
23. f(x) = (x2 – 1) / (x3 – 1) is undefined at x = 1 the value of f(x) at x = 1 such that it is
continuous at x = 1 is
(a) 3/2 (b) 2/3 (c) – 3/2 (d) none of these.
24. f(x) = 2x – |x| is
(a) undefined at x = 0 (b) discontinuous at x = 0
(c) continuous at x = 0 (d) none of these.
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:12)(cid:26)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:2)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:6)(cid:10)(cid:11)(cid:8)(cid:4)(cid:2)(cid:8)(cid:12)(cid:2)(cid:4)(cid:13)(cid:14)(cid:2)(cid:8)(cid:4)(cid:12)(cid:2)(cid:4)(cid:2)(cid:15)(cid:16)(cid:6)(cid:7)(cid:17)(cid:17)(cid:18)(cid:11)(cid:7)(cid:10)(cid:19)
25. If f(x) = 3, when x <2
f(x) = kx2, when x >2 is continuous at x = 2, then the value of k is
(a) ¾ (b) 4/3 (c) 1/3 (d) none of these.
x2-3x+2
≠
26. f(x) = x 1 becomes continuous at x = 1. Then the value of f(1) is
x-1
(a) 1 (b) –1 (c) 0 (d) none of these.
(x2-2x-3)
≠
27. f(x) = x –1 and f(x) = k, when x = –1 If(x) is continuous at x= –1 .
(x+1)
The value of k will be
(a) –1 (b) 1 (c) –4 (d) none of these.
⎛ ⎞
⎜x2- x⎟
28. l x i → m 1 ⎜ ⎜⎜⎝ x-1 ⎠ ⎟ ⎟ ⎟ is equal to
(a) 3 (b) –3 (c) 1/3 (d) none of these.
ex2
-1
29. lim is evaluated to be
x→0 x2
(a) 1 (b) ½ (c) –1 (d) none of these.
xn-2n
lim
30. If = 80 and n is a positve integer, then
x→2 x-2
(a) n = 5 (b) n = 4 (c) n = 0 (d) none of these.
x5/2-25/4
31. lim is equal to
x→ 2
x-21/4
(a) 1/ 10 (b) 10 (c) 20 (d) none of these.
⎛ ⎞
⎜ 1 x ⎟
32.
x
l
⎯
im
⎯→1
⎜⎜⎝
x2+x-2
-
x3-1
⎟ ⎟⎠ is evaluated to be
(a) 1/9 (b) 9 (c) – 1/9 (d) none of these.
⎡ ⎤
1 1 1 1
lim ⎢ + + + ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ + ⎥
33. n→∞ ⎢⎣6 62 63 6n⎥⎦ is
(a) 1/5 (b) 1/6 (c) – 1/5 (d) none of these.
(cid:10)(cid:11)(cid:12)(cid:12) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
34. The value of lim ux + vx + wx – 3 / x is
x→0
(a) uvw (b) log uvw (c) log (1/uvw) (d) none of these.
x
35. lim is equal to
log(1+x)
x→0
(a) 1 (b) 2 (c) –0.5 (d) none of these.
AAAAANNNNNSSSSSWWWWWEEEEERRRRRSSSSS
EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 88888(((((AAAAA)))))
1. a 2. b 3. c 4. a 5. c 6. a 7. b 8. c
9. b 10. a
EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 88888(((((BBBBB)))))
1. a 2. c 3. b 4. c 5. c 6. b 7. c 8. a
9. b 10. c 11. a 12. d 13. c 14. a 15. b 16. c
17. a 18. c 19. a 20. b 21. b 22. a 23. c 24. a
25. b 26. a 27. a 28. b 29. a 30. d 31. a 32. c
33. a 34. c 35. a 36. b 37. a
EEEEExxxxxeeeeerrrrrccccciiiiissssseeeee 88888(((((CCCCC)))))
1. a 2. bc 3. a,c 4. a 5. a 6. b 7. c 8. d
9. b 10. a 11. b 12. c 13. b 14. a 15. b 16. a
17. b 18. c 19. a 20. b 21. c 22. a 23. b 24. c
25. a 26. b 27. c 28. a 29. a 30. a 31. b 32. c
33. a 34. b 35. a
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:12)(cid:20)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:2)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:6)(cid:10)(cid:11)(cid:8)(cid:4)(cid:2)(cid:8)(cid:12)(cid:2)(cid:4)(cid:13)(cid:14)(cid:2)(cid:8)(cid:4)(cid:12)(cid:2)(cid:4)(cid:2)(cid:15)(cid:16)(cid:6)(cid:7)(cid:17)(cid:17)(cid:18)(cid:11)(cid:7)(cid:10)(cid:19)
AAAAADDDDDDDDDDIIIIITTTTTIIIIIOOOOONNNNNAAAAALLLLL QQQQQUUUUUEEEEESSSSSTTTTTIIIIIOOOOONNNNN BBBBBAAAAANNNNNKKKKK
1. The value of the limit when n tends to infinity of the expression
(7n3-8n2+10n-7)÷(8n3-9n2+5)is
(A) 7/8 (B) 8/7 (C) 1 (D) None
2. The value of the limit when n tends to infinity of the expression (n4-7n2+9)÷(3n2+5) is
(A) 0 (B) 1 (C) –1 (D) ∝
3. The value of the limit when n trends to infinity of the expression
(3n3+7n2-11n+19)÷(17n4+18n3-20n+45) is
(A) 0 (B) 1 (C) –1 (D) 1/ 2
4. The value of the limit when n tends to infinity of the expression (2n)÷[(2n-1)(3n+5)] is
(A) 0 (B) 1 (C) –1 (D) 1/ 2
5. The value of the limit when n tends to infinity of the expression
n1/3(n2+1)1/3(2n2+3n+1)-1/2 is
(A) 0 (B) 1 (C) –1 (D) 1/ 2
6. The value of the limit when x tends to a of the expression (xn-an)÷(x-a) is
(A) nan-1 (B) nan (C) (n-1)an-1 (D) (n+1)an+1
7. The value of the limit when x tends to zero of the expression (1+n)1/n is
(A) e (B) 0 (C) 1 (D) –1
( 1 )n
8. The value of the limit when n tends to infinity of the expression 1+ is
n
(A) e (B) 0 (C) 1 (D) –1
9. The value of the limit when x tends to zero of the expression [(1+x)n -1]÷x is
(A) n (B) n + 1 (C) n – 1 (D) n(n – 1)
10. The value of the limit when x tends to zero of the expression (ex–1)/x is
(A) 1 (B) 0 (C) – 1 (D) indeterminate
11. The value of the limit when x tends to 3 of the expression (x2+2x-15)/(x2-9) is
(A) 4/3 (B) 3/4 (C) 1/2 (D) indeterminate
(cid:10)(cid:11)(cid:12)(cid:21) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
Copyright -The Institute of Chartered Accountants of India
12. The value of the limit when x tends to zero of the expression [(a +x2)1/2-(a-x2)1/2]÷x2 is
(A) a-1/2 (B) a1/2 (C) a (D) a-1
13. The value of the limit when x tends to unity of the expression [(3+x)1/2-(5-x)1/2]÷(x2-1) is
(A) 1/4 (B) ½ (C) –1/4 (D) –1/2
14. The value of the limit when x tends to 2 of the expression (x-2)-1-(x2-3x+2)-1 is
(A) 1 (B) 0 (C) –1 (D) None
15. The value of the limit when n tends to infinity of the expression
2-n(n2+5n+6)[(n+4)(n+5)]-1 is
(A) 1 (B) 0 (C) –1 (D) None
lim n+1 1
16. The value of ÷
n →∝ n2 n
(A) 1 (B) 0 (C) –1 (D) None
lim
[n1/2+(n+1)1/2]-1÷n-1/2
17. Find →∝
n
(A) 1/2 (B) 0 (C) 1 (D) None
lim
(2n-1)(2n)n2(2n+1)-2(2n+2)-2
18. Find →∝
n
(A) 1/4 (B) 1/2 (C) 1 (D) None
lim
[(n3+1)1/2-n3/2]÷n3/2
19. Find →∝
n
(A) 1/4 (B) 0 (C) 1 (D) None
lim
[(n4+1)1/2-(n4-1)1/2]÷n-2
20. Find →∝
n
(A) 1/4 (B) 1/2 (C) 1 (D) None
lim
(2n-2)(2n+1)-1
21. Find →∝
n
(A) 1/4 (B) 1/2 (C) 1 (D) None
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:12)(cid:22)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:2)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:6)(cid:10)(cid:11)(cid:8)(cid:4)(cid:2)(cid:8)(cid:12)(cid:2)(cid:4)(cid:13)(cid:14)(cid:2)(cid:8)(cid:4)(cid:12)(cid:2)(cid:4)(cid:2)(cid:15)(cid:16)(cid:6)(cid:7)(cid:17)(cid:17)(cid:18)(cid:11)(cid:7)(cid:10)(cid:19)
lim
nn(n+1)-n-1÷n-1
22. Find →∝
n
(A) e-1 (B) e (C) 1 (D) –1
lim
(2n-1)2n(2n+1)-121-n
23. Find →∝
n
(A) 2 (B) 1/2 (C) 1 (D) None
lim
2n-1(10+n)(9+n)-12-n
24. Find →∝
n
(A) 2 (B) 1/2 (C) 1 (D) None
lim
[n(n+2)]÷(n+1)2
25. Find →∝
n
(A) 2 (B) 1/2 (C) 1 (D) None
lim
[n!3n+1]÷[3n(n+1)!]
26. Find →∝
n
(A) 0 (B) 1 (C) –1 (D) 2
lim
(n3+a)[(n+1)3a]-1(2n+1+a)(2n+a)-1
27. Find →∝
n
(A) 0 (B) 1 (C) –1 (D) 2
lim
(n2+1)[(n+1)2+1]-15n+15-n
28. Find →∝
n
(A) 5 (B) e-1 (C) 0 (D) None
lim
[nn.(n+1)!]÷[n!(n+1)n+1]
29. Find →∝
n
(A) 5 (B) e-1 (C) 0 (D) None
lim
[{1.3.5....(2n-1)}(n+1)4]÷[n4{1.3.5....(2n-1)(2n+1)}]
30. Find →∝
n
(A) 5 (B) e-1 (C) 0 (D) None
lim
[xn.(n+1)]÷[nxn+1]
31. Find →∝
n
(A) x-1 (B) x (C) 1 (D) None
(cid:10)(cid:11)(cid:12)(cid:23) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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lim
nn(1+n)-n
32. Find →∝
n
(A) e-1 (B) e (C) 1 (D) –1
lim
[(n+1)n+1.n-n-1-(n+1).n-1]-n
33. Find →∝
n
(A) (e-1)-1 (B) (e+1)-1 (C) e-1 (D) e+1
lim
(1+n-1)[1+(2n)-1]-1
34. Find →∝
n
(A) 1/2 (B) 3/2 (C) 1 (D) –1
lim
[4n2+6n+2]÷4n2
35. Find →∝
n
(A) 1/2 (B) 3/2 (C) 1 (D) –1
36. 3x2+2x-1 is continuous
(A) at x = 2 (B) for every value of x
(C) both (A) and (B) (D) None
x
(cid:2)
37. f(x) = , when x 0, then f(x) is
x
(A) discontinuous at x = 0 (B) continuous at x = 0
(C) maxima at x = 0 (D) minima at x = 0
38. e-1/x[1+e1/x]-1 is
(A) discontinuous at x = 0 (B) continuous at x = 0
(C) maxima at x = 0 (D) minima at x = 0
39. If f(x)=(x2-4)÷(x-2) for x<2, f(x)=4 for x=2 and f(x)=2 for x>2, then f(x) at x = 2 is
(A) discontinuous (B) continuous
(C) maxima (D) minima
≤
40. If f(x)=x for 0 x<1/2, f(x)=1 for x=1/2 and f(x)=1-x for 1/2<x<1 then at the
function is
(A) discontinuous (B) continuous
(C) left-hand limit coincides with f(1/2) (D) right-hand limit coincides with left-hand limit.
(cid:14)(cid:3)(cid:5)(cid:2)(cid:19) (cid:10)(cid:11)(cid:12)(cid:24)
Copyright -The Institute of Chartered Accountants of India
(cid:1)(cid:2)(cid:3)(cid:2)(cid:4)(cid:5)(cid:6)(cid:7)(cid:8)(cid:9)(cid:6)(cid:10)(cid:11)(cid:8)(cid:4)(cid:2)(cid:8)(cid:12)(cid:2)(cid:4)(cid:13)(cid:14)(cid:2)(cid:8)(cid:4)(cid:12)(cid:2)(cid:4)(cid:2)(cid:15)(cid:16)(cid:6)(cid:7)(cid:17)(cid:17)(cid:18)(cid:11)(cid:7)(cid:10)(cid:19)
41. If f(x)=9x÷(x+2) for x<1, f(1)=3, f(x)=(x+3)x-1 for x>1, then in the interval (–3, 3) the
function is
(A) continuous at x = –2
(B) continuous at x = 1
(C) discontinuous for values of x other than –2 1 in the interval
(D) None
AAAAANNNNNSSSSSWWWWWEEEEERRRRRSSSSS
1) A 2) D 3) A 4) A 5) D 6) A
7) A 8) A 9) A 10) A 11) A 12) A
13) A 14) A 15) B 16) A 17) A 18) A
19) B 20) C 21) C 22) A 23) A 24) B
25) C 26) A 27) D 28) A 29) B 30) C
31) A 32) A 33) A 34) A 35) C 36) C
37) A 38) A 39) A 40) A 41) D
(cid:10)(cid:11)(cid:12)(cid:10) (cid:1)(cid:13)(cid:14)(cid:14)(cid:13)(cid:15)(cid:8) (cid:4)(cid:7)(cid:13)(cid:16)(cid:17)(cid:1)(cid:17)(cid:6)(cid:15)(cid:1)(cid:18)(cid:8) (cid:5)(cid:6)(cid:19)(cid:5)
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