450 + Interview Questions in LIMITS IN GENERAL AWARENESS Page 3 InterviewSolution

101.	Differentiate the following with respect to xy = (√x + 1/√x)2
Answer» y = x + 1/x + 2 = x + x^-1 + 2 dy/dx = 1 - x^-2 + 0 = 1 - 1/x²

101.

Differentiate the following with respect to xy = (√x + 1/√x)2

Answer»

y = x + 1/x + 2 = x + x^-1 + 2

dy/dx = 1 - x^-2 + 0 = 1 - 1/x²

Discussion

102.	Differentiate the following with respect to xy = (x + 1/x)2, x ≠0
Answer» y = x² + 1/x² + 2 = x² + x^-2 + 2 dy/dx = 2x - 2x^-3 + 0 = 2x - 2x^-3

102.

Differentiate the following with respect to xy = (x + 1/x)2, x ≠0

Answer»

y = x² + 1/x² + 2

= x² + x^-2 + 2

dy/dx = 2x - 2x^-3 + 0

= 2x - 2x^-3

Discussion

103.	Explain the meaning of x → a .
Answer» Let x be a variable and ‘a’ be a constant. Since ‘x’ is a variable we can change its value at pleasure. It can be changed so that its value comes nearer and nearer to a. Then we say that x approaches ‘a’ and it is denoted by x → a .

103.

Explain the meaning of x → a .

Answer»

Let x be a variable and ‘a’ be a constant. Since ‘x’ is a variable we can change its value at pleasure. It can be changed so that its value comes nearer and nearer to a.

Then we say that x approaches ‘a’ and it is denoted by x → a .

Discussion

104.	Evaluate: \(\lim\limits_{x \to 0}(\frac{3^{2x}-2^{3x}}{x})\)
Answer» \(\lim\limits_{x \to 0}(\frac{3^{2x}-2^{3x}}{x})\) \(=\lim\limits_{x \to 0}\{\frac{(3^{2x}-1)-(2^{3x}-1)}{x}\}\) \(=\lim\limits_{x \to 0}\frac{3^{2x}-1}{x}-\lim\limits_{x \to 0}\frac{2^{3x}-1}{x}\) \(=\lim\limits_{x \to 0}\frac{1}{2}[\frac{3^{2x}-1}{x}]-3.[\lim\limits_{x \to 0}\frac{2^{3x}-1}{x}]\) = 2 ∙ (log 3) − 3(log 2) = log3² − log2³ = log \(\frac{9}{8}\)

104.

Evaluate: \(\lim\limits_{x \to 0}(\frac{3^{2x}-2^{3x}}{x})\)

Answer»

\(\lim\limits_{x \to 0}(\frac{3^{2x}-2^{3x}}{x})\)

\(=\lim\limits_{x \to 0}\{\frac{(3^{2x}-1)-(2^{3x}-1)}{x}\}\)

\(=\lim\limits_{x \to 0}\frac{3^{2x}-1}{x}-\lim\limits_{x \to 0}\frac{2^{3x}-1}{x}\)

\(=\lim\limits_{x \to 0}\frac{1}{2}[\frac{3^{2x}-1}{x}]-3.[\lim\limits_{x \to 0}\frac{2^{3x}-1}{x}]\)

= 2 ∙ (log 3) − 3(log 2)

= log3² − log2³

= log \(\frac{9}{8}\)

Discussion

105.	Define a derivative of f(x) at a point.
Answer» Suppose f is a real valued function and ‘a’ is a point in its domain of definition. The derivative of f at ‘a’ is defined by lim_(h→0) (f(a + h) - f(a))/h Provided this limit exists. Derivative of f(x) at 'a' is denoted by f'(a) ∴ f'(a) = lim_(h→a) (f(a + h) - f(a))/h

105.

Define a derivative of f(x) at a point.

Answer»

Suppose f is a real valued function and ‘a’ is a point in its domain of definition. The derivative of f at ‘a’ is defined by

lim_(h→0) (f(a + h) - f(a))/h

Provided this limit exists. Derivative of f(x) at 'a' is denoted by f'(a)

∴ f'(a) = lim_(h→a) (f(a + h) - f(a))/h

Discussion

106.	Find lim(x→4) (4x + 3)/(x - 2)
Answer» lim_(x →4) (4x + 3)/(x - 2) = (4(4) + 3)/(4 - 2) = 19/2

Discussion

107.	Find lim(x→1) πr2
Answer» lim_(x→1) πr² = (1)² = π

Discussion

108.	`lim_(xrarr oo)(1+sin.(1)/(n))^n` equalsA. `e^a`B. `e`C. `e^2a`D. `0`
Answer» Correct Answer - A We have, `lim_(xto oo) (1+sin(a)/(n))^n` `=e^(lim_(ntooo) nsin.(a)/(n))=e^(lim_(ntooo)sin((a//a)/(a//n))xx a )=e^(1xxa)=e^(a)`

Discussion

109.	Find lim(x→0) ((x + 1)2 - 1)/x
Answer» lim_(x→0) ((x + 1)² - 1)/x = lim_(x →0) (x² + 2x + 1 - 1)/x = lim_(x →0) (x + 2) = 0 + 2 = 2

109.

Find lim(x→0) ((x + 1)2 - 1)/x

Answer»

lim_(x→0) ((x + 1)² - 1)/x

= lim_(x →0) (x² + 2x + 1 - 1)/x

= lim_(x →0) (x + 2) = 0 + 2 = 2

Discussion

110.	Find lim(x →-1) (x10 + x5 + 1)/(x - 1)
Answer» lim_(x →-1) (x¹⁰ + x⁵ + 1)/(x - 1) = ((-1)¹⁰ + (-1)⁵ + 1)/(-1 - 1) = (1 - 1 + 1)/-2 = - 1/2

110.

Find lim(x →-1) (x10 + x5 + 1)/(x - 1)

Answer»

lim_(x →-1) (x¹⁰ + x⁵ + 1)/(x - 1)

= ((-1)¹⁰ + (-1)⁵ + 1)/(-1 - 1)

= (1 - 1 + 1)/-2 = - 1/2

Discussion

111.	Find lim(x →a) (x√x - a√a)/(x - a)
Answer» lim_(x →a) (x√x - a√a)/(x - a) = lim_(x →a) (x^3/2 - a^3/2)/(x - a) = 3/2a^{3/2 - 1} = 3/2a^1/2

111.

Find lim(x →a) (x√x - a√a)/(x - a)

Answer»

lim_(x →a) (x√x - a√a)/(x - a) = lim_(x →a) (x^3/2 - a^3/2)/(x - a)

= 3/2a^{3/2 - 1} = 3/2a^1/2

Discussion

112.	Find lim(2x →-1) (8x3 + 1)/(2x + 1)
Answer» lim_(2x →-1) (8x³ + 1)/(2x + 1) ((2x)³ - (-1)³)/(2x - (-1)) = 3(-1)² = 3

112.

Find lim(2x →-1) (8x3 + 1)/(2x + 1)

Answer»

lim_(2x →-1) (8x³ + 1)/(2x + 1)

((2x)³ - (-1)³)/(2x - (-1))

= 3(-1)² = 3

Discussion

113.	\(\lim\limits_{x \to 0}(\frac{\sqrt {1+x^2}-\sqrt {1+x}}{\sqrt {1+x^3}-\sqrt {1+x}})= ?\)
Answer» \(\lim\limits_{x \to 0}(\frac{\sqrt {1+x^2}-\sqrt {1+x}}{\sqrt {1+x^3}-\sqrt {1+x}})\) \(=\lim\limits_{x \to 0}\{\frac{\sqrt {1+x^2}-\sqrt {1+x}}{\sqrt {1+x^3}-\sqrt {1+x}}\times\)\(\frac{\sqrt {1+x^3}+\sqrt {1+x}}{\sqrt {1+x^3}+\sqrt {1+x}}\times\)\(\frac{\sqrt {1+x^2}+\sqrt {1+x}}{\sqrt {1+x^2}+\sqrt {1+x}}\}\) \(=\lim\limits_{x \to 0}\{\frac{(1+x^2)- (1+x)}{(1+x^3)-(1+x)}\times\)\(\frac{\sqrt{1+x^3}+ \sqrt {1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}\}\) \(=\lim\limits_{x \to 0}\{\frac{x(x-1)}{(x^3-x)}\times\)\(\frac{\sqrt{1+x^3}+ \sqrt {1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}\}\) \(=\lim\limits_{x \to 0}\frac{x(x-1)(\sqrt{1+x^3}+\sqrt{1+x})}{x(x-1)x(x+1)(\sqrt{1+x^2}+\sqrt{1+x})}\) \(=\lim\limits_{x \to 0}\frac{(\sqrt{1+x^3}+\sqrt{1+x})}{(x+1)(\sqrt{1+x^2}+\sqrt{1+x})}\) \(=\frac{(\sqrt{1+0^3}+\sqrt{1+0})}{(0+1)(\sqrt{1+0^2}+\sqrt{1+0})}\) \(= \frac{2}{1.2}\) = 1

113.

\(\lim\limits_{x \to 0}(\frac{\sqrt {1+x^2}-\sqrt {1+x}}{\sqrt {1+x^3}-\sqrt {1+x}})= ?\)

Answer»

\(\lim\limits_{x \to 0}(\frac{\sqrt {1+x^2}-\sqrt {1+x}}{\sqrt {1+x^3}-\sqrt {1+x}})\)

\(=\lim\limits_{x \to 0}\{\frac{\sqrt {1+x^2}-\sqrt {1+x}}{\sqrt {1+x^3}-\sqrt {1+x}}\times\)\(\frac{\sqrt {1+x^3}+\sqrt {1+x}}{\sqrt {1+x^3}+\sqrt {1+x}}\times\)\(\frac{\sqrt {1+x^2}+\sqrt {1+x}}{\sqrt {1+x^2}+\sqrt {1+x}}\}\)

\(=\lim\limits_{x \to 0}\{\frac{(1+x^2)- (1+x)}{(1+x^3)-(1+x)}\times\)\(\frac{\sqrt{1+x^3}+ \sqrt {1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}\}\)

\(=\lim\limits_{x \to 0}\{\frac{x(x-1)}{(x^3-x)}\times\)\(\frac{\sqrt{1+x^3}+ \sqrt {1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}\}\)

\(=\lim\limits_{x \to 0}\frac{x(x-1)(\sqrt{1+x^3}+\sqrt{1+x})}{x(x-1)x(x+1)(\sqrt{1+x^2}+\sqrt{1+x})}\)

\(=\lim\limits_{x \to 0}\frac{(\sqrt{1+x^3}+\sqrt{1+x})}{(x+1)(\sqrt{1+x^2}+\sqrt{1+x})}\)

\(=\frac{(\sqrt{1+0^3}+\sqrt{1+0})}{(0+1)(\sqrt{1+0^2}+\sqrt{1+0})}\)

\(= \frac{2}{1.2}\)

= 1

Discussion

114.	Find the value of 'a' and 'b' if \(\lim\limits_{x \to 2}f(x)\) and \(\lim\limits_{x \to 4}f(x)\)exists where.\(f(x) = \begin{cases} x^2+as+b & \quad \text, 0\leq{x} <{2}\\ 3x+2, & \quad \text, 2\leq{x}\leq{4}\\2ax+5b & \quad \text,4<x\leq{8} \end{cases} \)
Answer» Given, \(f(x) = \begin{cases} x^2+as+b & \quad \text, 0\leq{x} <{2}\\ 3x+2, & \quad \text, 2\leq{x}\leq{4}\\2ax+5b & \quad \text,4<x\leq{8} \end{cases} \) To find \(\lim\limits_{x \to 2}f(x)\) L.H.L \(=\lim\limits_{x \to 2^-}f(x)\) \(=\lim\limits_{x \to 2^-}(x^2+ax+b)\) = 2²+ a∙2 + b = 2a + b + 4 And, R.H.L \(=\lim\limits_{x \to 2^+}f(x)\) \(=\lim\limits_{x \to 2^+}(3x+2)\) = 3 ∙ 2 + 2 = 8 Since \(\lim\limits_{x \to 2}f(x) \) exists, ∴ \(\lim\limits_{x \to 2^-}f(x)\) \(=\lim\limits_{x \to 2^+}f(x)\) ⇒ 2a + b + 4 = 8 ⇒ 2a + b = 4 … (1) To find \(\lim\limits_{x \to 2}f(x)\). L.H.L \(=\lim\limits_{x \to 4^-}f(x)\) \(=\lim\limits_{x \to 4^-}(3x+2)\) = 3 ∙ 4 + 2 = 14 And R.H.L \(=\lim\limits_{x \to 4^+}f(x)\) \(=\lim\limits_{x \to 4^+}(2ax+5b)\) = 2a. 4 + 5b = 8a + 5b Since \(\lim\limits_{x \to 4}f(x) \) exists. ∴ \(\lim\limits_{x \to 4^-}f(x) \) = \(\lim\limits_{x \to 4^+}f(x) \) ⇒ 8a + 5b = 14 … (2) From (1) and (2), a = 3 and b = -2.

114.

Find the value of 'a' and 'b' if \(\lim\limits_{x \to 2}f(x)\) and \(\lim\limits_{x \to 4}f(x)\)exists where.\(f(x) = \begin{cases} x^2+as+b & \quad \text, 0\leq{x} <{2}\\ 3x+2, & \quad \text, 2\leq{x}\leq{4}\\2ax+5b & \quad \text,4<x\leq{8} \end{cases} \)

Answer»

Given,

\(f(x) = \begin{cases} x^2+as+b & \quad \text, 0\leq{x} <{2}\\ 3x+2, & \quad \text, 2\leq{x}\leq{4}\\2ax+5b & \quad \text,4<x\leq{8} \end{cases} \)

To find \(\lim\limits_{x \to 2}f(x)\)

L.H.L

\(=\lim\limits_{x \to 2^-}f(x)\)

\(=\lim\limits_{x \to 2^-}(x^2+ax+b)\)

= 2²+ a∙2 + b

= 2a + b + 4

And,

R.H.L

\(=\lim\limits_{x \to 2^+}f(x)\)

\(=\lim\limits_{x \to 2^+}(3x+2)\)

= 3 ∙ 2 + 2 = 8

Since \(\lim\limits_{x \to 2}f(x) \) exists,

∴ \(\lim\limits_{x \to 2^-}f(x)\) \(=\lim\limits_{x \to 2^+}f(x)\)

⇒ 2a + b + 4 = 8

⇒ 2a + b = 4 … (1)

To find \(\lim\limits_{x \to 2}f(x)\).

L.H.L

\(=\lim\limits_{x \to 4^-}f(x)\)

\(=\lim\limits_{x \to 4^-}(3x+2)\)

= 3 ∙ 4 + 2 = 14

And

R.H.L

\(=\lim\limits_{x \to 4^+}f(x)\)

\(=\lim\limits_{x \to 4^+}(2ax+5b)\)

= 2a. 4 + 5b

= 8a + 5b

Since \(\lim\limits_{x \to 4}f(x) \) exists.

∴ \(\lim\limits_{x \to 4^-}f(x) \) = \(\lim\limits_{x \to 4^+}f(x) \)

⇒ 8a + 5b = 14 … (2)

From (1) and (2),

a = 3 and b = -2.

Discussion

115.	If `lim_(x to 0) (cos4x+acos2x+b)/(x^(4))` is finite, find a and b using expansion formula. Also, find the limit.
Answer» `underset(xto0)lim(cos4x+acos2x+b)/(x^(4))="Finite"` Using expansion formula for `cos4x" and "cos2x,` we get `underset(xto0)lim((1-(4x)^(2)/(2!)+(4x)^(4)/(4!))+a(1-(2x)^(2)/(2!)+(2x)^(4)/(4!))+b)/(x^(4))` or `" "underset(xto0)lim((1+a+b)+(-8-2a)x^(2)+((32)/(3)+2/3a)x^(4)+...)/(x^(4))` or `" "1+a+b=0` `-8-2=0` Solving (1) and (2) for a and b, we get `a=-4" and " b=3` Also, Limit`=32/3+2/3a=(32-8)/(3)=8`

Discussion

116.	Evaluate `lim_(xto0) ((1)/(x^(2))-(1)/(sin^(2)x)).`
Answer» `underset(xto0)lim((1)/(x^(2))-(1)/(sin^(2)x))` `=underset(xto0)lim((sin^(2)x-x^(2))/(x^(2)sin^(2)x))` `=underset(xto0)lim((sinx+x)(sinx-x))/(x^(2)sin^(2)x)` `=underset(xto0)lim(((x-(x^(3))/(3!)+(x^(5))/(5!)...)+x)((x-(x^(3))/(3!)+(x^(5))/(5!)...)-x))/(x^(2)(x-(x^(3))/(3!)+(x^(5))/(5!)...)^(2))` `=underset(xto0)lim(-(2-(x^(2))/(3!)+(x^(4))/(5!)...)((1)/(3!)-(x^(2))/(5!)...))/(x^(2))=-1/3`

Discussion

117.	Let `kgt0 and lambda =lim_(xrarr0) (k(1-4sqrt(k^2-x^2)))/(x^2sqrt(k^2-x^2))` be finite. Then the value of `lambdak`, isA. `lambda=8,k=(1)/(2)`B. `lambda=8,k=(1)/(4)`C. `lambda=4, k=(1)/(2)`D. `6lambda=4,k=(1)/(4)`
Answer» Correct Answer - B It is given that ` lambda =lim_(xto0) (k(1-4sqrt(k^2-x^2)))/(x^2sqrt(k^2-x^2)) "exists "` `rArrlim_(xto0)k(1-4sqrt(k^2-x^2))=0 rArr 1-4k=0rArr k=(1)/(4)` `therefore lambda =lim_(xto0) (1-sqrt(1-16x^2))/(x^2sqrt(1-16x^2))` ` rArr lambda =lim_(xto0)(1)/(x^2(1+sqrt(1-16x^2)sqrt(1-16x^2))` ` rArr lambda =16lim_(xto0)(1)/(x^2(1+sqrt(1-16x^2))sqrt(1-16x^2))` Hence ,`lambda =8 and k=(1)/(4)`

Discussion

118.	If `f(x)=0`is a quadratic equation such that `f(-pi)=f(pi)=0`and `f(pi/2)=-(3pi^2)/4,`then `("lim")_(xvecpi)(f(x))/("sin"(sinx)`is equal to0 (b) `pi`(c)`2pi`(d)none of these
Answer» Correct Answer - C It is given that `f(-pi)=f(pi)=0`. Therefore ,`-pi` and `pi` are zeroes of `f(x)`. `therefore f(x)=a(x+pi)(x-pi),a ne 0` ` rArr f((pi)/(2))=-(3api^2)/(4)rArr(-3pi^2)/(4)=-(3api^2)/(4)rArr a=1` `therefore f(x)=(x+pi)(x-pi)` So, `lim_(xto-pi) (f(x))/(sin(sinx))` `=lim_(xto-pi)((x-pi)(x+pi))/((sin(sinx))/(sinx)xx sinx )=-lim_(xto-pi)((x-pi))/((sin(sinx))/(sinx))xx((x+pi))/(sin(pi+x))=-((-pi-pi))/(1)xx1=2pi`

Discussion

119.	Let `f(x)=[x]+[-x]`, where `[x]` denotes the greastest integer less than or equal to x . Then, for any integer mA. `lim_(xtom) f(x)=f(m)`B. `lim_(xtom)f(x)ne f(m)`C. `lim_(xtom)f(x)` does not existD. none of these
Answer» Correct Answer - B We have ` f(x)=[x]+[-x]` `rArr f(x)={([x]-[x]=0,"if x is an integer",),([x]+(-[x]-1)=-1,"if x is not an integer",):}` ` therefore lim_(xotm^-)f(x)=lim_(xtom^+)f(x)=-1and , f(m)=0`. Hence, `lim_(xtom^-)f(x)ne f(m)`.

Discussion

120.	The value of `lim_(xrarr oo) {(x^4sin((1)/(x))+x^2)/(1+\|x^3\|)}` isA. 1B. -1C. 0D. `oo`
Answer» Correct Answer - B We have , `lim_(xto oo) {(x^4sin((1)/(x))+x^2)/(1+\|x^3\|)}` ` =lim_(yto oo)(y^4sin((-1)/(y))+y^2)/(1+\|x\|^3),"where "y=-x` ` =lim_(yto oo) (-y^4sin((1)/(y))+y^2)/(1+y^3)` ` lim_(yto oo) (-ysin((1)/(y))+(1)/(y))/((1)/y^(3)+1)=(-1+0)/(0+1)=-1`

Discussion

121.	If `[.]` denotes the greatest intger function, then `lim_(xrarr0) (tan([-2pi^2]x^2)-x^2tan[-2pi^2])/(sin^2x)` is equal toA. `20 +tan 20`B. `20+tan 20`C. `20`D. none of these
Answer» Correct Answer - A We have, `[-2pi^2]=-20`. `therefore lim_(xto0) (tan([-2pi^2]x^2)-x^2tan[-2pi^2])/(sin^2x)` `=lim_(xto0) (tan(-20x^2)-x^2tan(-20))/(sin^2x)` `=lim_(xto0) -(tan20x^2)/(20x^2)xx20xx(x^2)/(sin^2x)xx((x)/(sinx))^2tan20 =-20+tan20`.

Discussion

122.	Find lim(x→1) (x3 - 1)/(x - 1)
Answer» lim_(x→1) (x³ - 1)/(x - 1) = lim_(x→1) (x³ - 1³)/(x - 1) = 3(1²) = 3

122.

Find lim(x→1) (x3 - 1)/(x - 1)

Answer»

lim_(x→1) (x³ - 1)/(x - 1) = lim_(x→1) (x³ - 1³)/(x - 1)

= 3(1²) = 3

Discussion

123.	Prove that \(\lim\limits_{x \to a^+}\) [x] = [a] for all a ∈ R. Also, prove that \(\lim\limits_{x \to 1^-}\)[x] = 0lim [x], x ∈ a+ = [a]
Answer» To Prove : lim [x], x ∈ a⁺= [a] L.H.S = \(\lim\limits_{x \to a^+}\) [x] = \(\lim\limits_{h \to 0}\)[a+h] = [a] (Since, [a + h] = [a]) Hence, Proved. Also, To prove : \(\lim\limits_{x \to 1^-}\) [x] = 0 L.H.S = \(\lim\limits_{x \to 1^-}\) [x] = \(\lim\limits_{h \to 0}\) [1-h] = 0 (Since, [1 – h] = 0) Hence, Proved.

123.

Prove that \(\lim\limits_{x \to a^+}\) [x] = [a] for all a ∈ R. Also, prove that \(\lim\limits_{x \to 1^-}\)[x] = 0lim [x], x ∈ a+ = [a]

Answer»

To Prove : lim [x], x ∈ a⁺= [a]

L.H.S = \(\lim\limits_{x \to a^+}\) [x] = \(\lim\limits_{h \to 0}\)[a+h] = [a]

(Since, [a + h] = [a])

Hence, Proved.

Also,

To prove : \(\lim\limits_{x \to 1^-}\) [x] = 0

L.H.S = \(\lim\limits_{x \to 1^-}\) [x]

= \(\lim\limits_{h \to 0}\) [1-h]

= 0

(Since, [1 – h] = 0)

Hence, Proved.

Discussion

124.	Evaluate `lim_(xtoa) (sqrt(a+2x)-sqrt(3x))/(sqrt(3a+x)-2sqrt(x)),(ane0).`
Answer» We have, `underset(xtoa)lim(sqrt(a+2x)-sqrt(3x))/(sqrt(3a+x)-2sqrt(x))" "("from"0/0)` `=underset(xtoa)lim((sqrt(a+2x)-sqrt(3x))(sqrt(a+2x)+sqrt(3x)))/((sqrt(3a+x)-2sqrt(x))(sqrt(3a+x)+2sqrt(x)))` `" "((sqrt(3a+x)+2sqrt(x)))/((sqrt(a+2x)+sqrt(3x)))" "("from"0/0)` `=underset(xtoa)lim((a+2x-3x))/((3a+x-4x))((sqrt(3a+x)+2sqrt(x)))/((sqrt(a+2x)+sqrt(3x)))` `=underset(xtoa)lim(sqrt(3a+x)+2sqrt(x))/(3(sqrt(a+2x)+sqrt(3x)))` `=(sqrt(3a+x)+2sqrt(x))/(3(sqrt(a+2a)+sqrt(3a)))=1/3.(4sqrt(a))/(2sqrt(3a))=(2)/(3sqrt(3))`

Discussion

125.	Evaluate`lim_(xto0) (2^(x)-1)/(sqrt(1+x)-1).`
Answer» We have `underset(xto0)lim(2^(x)-1)/(sqrt(1+x)-1)" "`(0/0 from) `=underset(xto0)lim(2^(x)-1)/(sqrt(1+x)-1)((sqrt(1+x)+1))/((sqrt(1+x)+1))` `=underset(xto0)lim(2^(x)-1)/(x)underset(xto0)lim(sqrt(1+x)+1)` =(log2)2 =log4

Discussion

126.	Evaluate `lim_(xto0) (sqrt(2+x)-sqrt(2))/(x).`
Answer» When `xto0,` the expression `underset(xto0)lim(sqrt(2+x)-sqrt(2))/(x).` takes the form 0/0. Rationalizing the numerator, we have `underset(xto0)lim(sqrt(2+x)-sqrt(2))/(x)=underset(xto0)lim((sqrt(2+x)-sqrt(2))(sqrt(2+x)+sqrt(2)))/(x(sqrt(2+x)+sqrt(2)))` `=underset(xto0)lim(2+x-2)/(x(sqrt(2+x)+sqrt2))` `=underset(xto0)lim(1)/(sqrt(2+x)+sqrt(2))=(1)/(2sqrt(2))`

Discussion

127.	Evaluate `lim_(ntooo) (4^(n)+5^(n))^(1//n)`
Answer» We have, `underset(ntooo)lim(4^(n)+5^(n))^(1//n)" "(oo^(0)"from")` `=underset(ntooo)lim5(1+((4)/(5))^(n))^(1//n)` `=5" "(because((4)/(5))^(n)to0" as n"tooo)`

Discussion

128.	Evaluate `lim_(xtooo) (log_(e)x)/(x)`
Answer» We have `underset(xtooo)lim(log_(e)x)/(x)` When `xtooo, log_(e)xtooo.` So, we have indeterminate form `(oo)/(oo).` But `log_(e)x` is very sluggish function, i.e., increases very slowly compared to x. So, at ifinity though both `log_(e)x` and x approaches to infinity but the cifference between these infinity values is also infinity, i.e., x is far more infinite than `log_(e)x.` Therefore `underset(xtooo)lim(log_(e)x)/(x)=0`

Discussion

129.	In the neighbourhood of `x=0` it is known that `1+\|x\|lt(e^(x)-1)/(x)lt1-\|x\|"then find"lim_(xto0)(e^(x)-1)/(x).`
Answer» We have `1+\|x\|lt(e^(x)1-)/(x)lt1-\|x\|"for "x in(0-delta,0+delta).` `:." "`Using Sandwich rule, `underset(xto0)lim(1+\|x\|)ltunderset(xto0)lim(e^(x)-1)/(x)ltunderset(xto0)lim(1-\|x\|)` `implies" "1ltunderset(xto0)lim(e^(x)-1)/(x)lt1` `underset(xto0)lim(e^(x)-1)/(x)=1`

Discussion

130.	If `0ltlog_(e)xltsqrt(x)` for all `xlt1,` then find the value of `lim_(xtooo) (log_(e)x)/(x).`
Answer» As `0ltlog_(e)xltsqrt(x)(xlt1)` `implies" "0ltunderset(xtooo)lim(log_(e)x)/(x)lt(1)/(sqrt(x)` `implies" "oltunderset(xtooo)lim(log_ex)/(x)ltunderset(xtooo)lim(1)/(sqrt(x))` `implies" "0ltunderset(xtooo)lim(log_(e)x)/(x)lt0` `underset(xtooo)lim(log_(x)x)/(x)=0" "`(Using Sandwich Rule)

Discussion

131.	If `f(x)=(2)/(x-3),g(x)=(x-3)/(x+4)," and "h(x)=-(2(2x+1))/(x^(2)+x-12)," then "lim_(xto3) [f(x)+g(x)+h(x)]" is "`A. 1B. `oo`C. `sqrt(2)`D. none of these
Answer» Correct Answer - C We have `f(x)+g(x)+h(x)=(x^(2)-4x+17-4x-2)/(x^(2)+x-12)` `=(x^(2)-8x+15)/(x^(2)+x-12)=((x-3)(x-5))/((x-3)(x+4))` `:. underset(xto3)lim[f(x)+g(x)+h(x)]=underset(xto3)lim((x-3)(x-5))/((x-3)(x+4))=-(2)/(7)`

Discussion

132.	Find the limits of the following: `(i)lim_(xto0) (sin3x)/(x)" "(ii)lim_(xto0) (sin7x)/(sin4x)" "(iii)lim_(xto0) (1-cos^(2)x)/(x^(2))`
Answer» `(i)underset(xto0)lim(sin3x)/(x)=3underset(xto0)lim(sin3x)/(3x)` `3xx1["as " underset(xto0)lim(sin(f(x)))/(f(x))=1, if underset(xto0)limf(x)=0]` =3 `(ii)underset(xto0)lim(sin7x)/(sin4x)=underset(xto0)lim((sin7x)/(7x))/((sin4x)/(4x))xx(7x)/(4x)` `(underset(xto0)lim((sin7x))/(7x))/(underset(xto0)lim(sin4x)/(4x))xx7/4` `=1/1xx7/4=7/4` `(iii)underset(xto0)lim(1-cos2x)/(x^(2))=underset(xto0)lim(2sin^(2)x)/(x^(2))` `=2(underset(xto0)lim(sinx)/(x))^(2)=2`

Discussion

133.	If `lim_(xto0) (x^(-3)sin3x+ax^(-2)+b)` exists and is equal to 0, thenA. `a=1`B. `a=0`C. `b=1`D. `b=-1`
Answer» Correct Answer - A `underset(xto0)lim((sin3x)/(x^(3))+(a)/(x^(2))+b)=underset(xto0)lim(sin3x+ax+bx^(3))/(x^(3))` `=underset(xto0)lim(3(sin3x)/(3x)+a+bx^(2))/(x^(2))` For existence, `(3+a)=0` or `a=-3` `:. L=underset(xto0)lim(sin3x-3x+bx^(3))/(x^(3))` `=27underset(t to0)lim(sint-t)/(t^(3))+b=0(3x=t)` `=-(27)/(6)+b=0` or `b=(9)/(2)`

Discussion

134.	The value of `lim_(x to oo ) (log_(e)(log_(e)x))/(e^(sqrt(x)))` is _________.A. `(pi)/(2)`B. 1C. `-pi`D. `pi`
Answer» Correct Answer - `(0)` `Let L=underset(xtooo)lim(log_(e)(log_(e)x))/(e^(sqrt(x)))" "((oo)/(oo)" form")` `=underset(xtooo)lim((1)/(xlog_(e)x))/(e^(sqrt(x)(1)/(2sqrt(x))))=underset(xtooo)lim(2sqrt(x))/(e^(sqrt(x))xlog_(e)x)` `=underset(xtooo)lim(2)/(e^(sqrt(x)sqrt(x)log_(e)x))` `=0`

Discussion

135.	The value of `lim_(x to oo ) (x-x^(2)log_(e)(1+(1)/(x)))` is _______.A. 1B. `(1)/(2)`C. `(1)/(4)`D. 2
Answer» Correct Answer - `(0.5)` Let` x=1//y` `:." "underset(xtooo)lim(x-x^(2)log_(e)(1+(1)/(x)))=underset(yto0)lim((1)/(y)-(log_(e)(1+y))/(y^(2)))` `=underset(yto0)lim((y-log_(e)(1+y))/(y^(2)))` `=underset(yto0)lim((y-(y-(y^(2))/(2)))/(y^(2)))=1//2`

Discussion

136.	Differentiate the following with respect to x(i) y = a(x - 2)(x - b)(ii) y = (ax2 + b)2
Answer» (i) y = x²- (a + b)x + ab ∴ dy/dx = 2x - (a + b) + 0 = 2x - (a + b) (ii) y = (ax² + b)² = a²x⁴ + 2abx² + b² ∴ dy/dx = a²(4x³) + 2ab(2x) + 0 = 4a²x³ + 4abx

136.

Differentiate the following with respect to x(i) y = a(x - 2)(x - b)(ii) y = (ax2 + b)2

Answer»

(i) y = x²- (a + b)x + ab

∴ dy/dx = 2x - (a + b) + 0

= 2x - (a + b)

(ii) y = (ax² + b)²

= a²x⁴ + 2abx² + b²

∴ dy/dx = a²(4x³) + 2ab(2x) + 0

= 4a²x³ + 4abx

Discussion

137.	Evaluate `lim_(ntooo) (1)/(n^(2(log_(e)n-log_(e)(n-1)))+n)`.
Answer» `underset(ntooo)lim(1)/(n^(2(log_(e)n-log_(e)(n-1)))+n)` `=underset(ntooo)lim(1)/(n^(2)("log"_(e)(n)/(n+1))+n)` `=underset(ntooo)lim(1)/(n^(2)("log"_(e)(1)/(1+(1)/(n)))+n)` `=underset(xto0)lim(1)/((1)/(x^(2))("log"_(e)(1)/(1+x))+(1)/(x))` `=underset(xto0)lim(1)/(-(log_(e)(1+x))/(x^(2))+(1)/(x))` `=underset(xto0)lim(x^(2))/(x-log_(e)(1+x))` `=underset(xto0)lim(x^(2))/(x-(x-(x^(2))/(2)))=2`

Discussion

138.	If `lim_(xrarroo) (8x^3+mx^2)^(1//3)-nx` exists and is equal to 1 , then the vlaue of `(m)/(n)` isA. `(1)/(6)`B. 6C. 3D. `(1)/(3)`
Answer» Correct Answer - B It is given that `lim_(xtooo) {(8x^3+mx^2)^(1//3)-nx}` exists and is equal to 1 `rArr lim_(xto oo) {2x(1+(m)/(8x))^(1//3)-nx}` exists and is equal to 1 `rArr lim_(xtooo) {2x(1+(m)/(24x)-(m^2)/(576x^2)+......)-nx}` exists and is equal to 1 `rArr 2-n=0and, (m)/(12)=1` `rArr n=2 and m=12 rArr (m)/(n)=6`.

Discussion

139.	If ` a .lim_(xrarr 1) x^(1//1-x)+b=e^(-1)(a ge 1,bge 0)` , thenA. `a=1,b=e^-1`B. `a=2,b=e^-1`C. `a=-1,b=e^-1`D. `a=1,b=0`
Answer» Correct Answer - D We have `a. lim_(xto1) x^(1//1-x+b=e^(-1))` ` rArr a. lim_(xto1)[1+(x-1)]^(1//1-x)+b=e^(-1)` ` rArr ae ^(lim_(xto1)(x-1)/(1-x))+b=e^(-1)` ` rArr a.e ^-1+b=e^-1rArr a=1and b=0`.

Discussion

140.	`lim_(xrarr1) (sinsqrt(x))/(sqrt(sinx))` is equal to
Answer» Correct Answer - B We have, `lim_(xto0^+) (sinsqrt(x))/(sqrt(sinx))=lim_(xto0^+)((sinsqrt(x))/(sqrt(x)))/(sqrt((sinx)/(x)))=(1)/(1)=1`.

Discussion

141.	`lim_(xrarr -1) (cos 2 -cos 2x)/(x^2-\|x\|)` is equaol toA. 2B. `sin 2`C. `2sin2`D. none of these
Answer» Correct Answer - C We have, `lim_(xto -1) (cos 2 -cos 2x)/(x^2-\|x\|)` ` =lim_(xto-1)(2sin(1+x)sin(x-1))/(x^2+x)[because \|x\|=-x " for " x lt 0]` ` =2 lim_(xto-1) (sin(xto1))/(xto1)xx(sin(x-1))/(x)` ` =2xx1 xx(sin(-2))/((-1))=2sin 2`.

Discussion

142.	`Lim_(x->oo) ((x/(x+1))^a + sin (1/x))^x` is equal toA. `e^a-1`B. `e^1-a`C. `e`D. `0`
Answer» Correct Answer - B We have `lim_(xto oo) {((x)/(x+1))^a a+sin.(1)/(x)}^x` `lim_(xto oo) {(1)/(x+(1)/(x))^a +sin.(1)/(x)}^x` ` lim_(xto oo) {1+(1+(1)/(x))^-a+ sin .(1)/(x)-1}^x` `= lim_(y to oo) {1+(1+y)^-a+siny 1}^(1//y)`, where `y-(1)/(x)` `=e^(lim_(xto0)((1+y)^-1+siny-1)/(y)` `=e^(lim_(yto0) {(siny)/(y)+((1+y)^-a)/(y)}` ` =e^(1-a)[because lim_(yto0) (siny)/(y)=1and lim_(yto0) ((1+y)^-a-1)/(y)=-a]`

Discussion

143.	If `{x}` denotes the fractional part of x, then `lim_(xrarr1) (x sin {x})/(x-1)`, is
Answer» Correct Answer - C We, ` lim_(xto1^-)(xsin{x})/(x-1)=lim_(xto1)(xsinx)/(x-1)to-oo` and ` lim_(xto1^+)(sin{x})/(x-1)=lim_(xto1^-) ((x sin x-1))/(x-1)=1xx1=1` Clearly , `lim_(xto1^-)(xsin{x})/(x-1) ne lim_(xto1^-)(xsin{x})/(x-1)` So `lim_(xto1^-)(xsin{x})/(x-1)` does not exist.

Discussion

144.	If {x} denotes the fractional part of x, then `underset(x to 0)(lim) ({x})/(tan {x})` is equal toA. 1B. 0C. -1D. none of these
Answer» Correct Answer - D We have ` lim_(xto0^-)({x})/(tan{x})` ` =lim_(xto0^-)(x-[x])/(tan (x-[x]))=lim_(xto0^-)(x+1)/(tan(x+1))=(1)/(tanx+1)=(1)/(tan1)=cot 1` and ` lim_(xto0^-)({x})/(tan{x})=lim_(xto0^+)(x-[x])/(tan(x-[x]))=lim_(xto0^+)(x)/(tanx)=1` Clearly , `lim_(xto0^-)({x})/(tan{x}) ne lim_(xt0^+)({x})/(tan{x})` So , `lim_(xto0) ({x})/(tan{x})` does not exist.

Discussion

145.	Evaluate: lim(x→0)(log(1 + x3))/sin3x
Answer» lim_(x→0)(log(1 + x³))/sin³x = lim_(x→0)(log(1 + x³))/x³ . x³/sin³x = 1 x 1 = 1

145.

Evaluate: lim(x→0)(log(1 + x3))/sin3x

Answer»

lim_(x→0)(log(1 + x³))/sin³x

= lim_(x→0)(log(1 + x³))/x³ . x³/sin³x

= 1 x 1 = 1

Discussion

146.	Discuss the limit of f(x) = x3 at x = 1.
Answer» LHL = lim_(x→1^-)x³ = 1 RHL = lim_(x →1⁺)x³ = 1 ⇒ LHL = RHL = 1 ∴ lim_(x→1) f(x) = 1

146.

Discuss the limit of f(x) = x3 at x = 1.

Answer»

LHL = lim_(x→1^-)x³ = 1

RHL = lim_(x →1⁺)x³ = 1

⇒ LHL = RHL = 1

∴ lim_(x→1) f(x) = 1

Discussion

147.	Evaluate `lim_(xto0) (e-(1+x)^(1//x))/(x)`.
Answer» `underset(xto0)lim(e-(1+x)^(1//x))/(x)=underset(xto0)lim(e-e^((1)/(x)log_(e)(1+x)))/(x)` `=underset(xto0)lim(e-e^((1)/(x)(x-(x^(3))/(2)+(x^(3))/(3)+(x^(4))/(4)+...)))/(x)` `=underset(xto0)lim(e-e.e^(-((x)/(2)+(x^(2))/(3)+(x^(3))/(4)+...)))/(x)` `=-e.underset(xto0)lim(e^(-((x)/(2)-(x^(2))/(3)+(x^(3))/(4)......))-1)/(-((x)/(2)-(x^(2))/(3)+(x^(3))/(4)...)).(-((x)/(2)-(x^(2))/(3)+(x^(3))/(4)...))/(x)` `=(e)/(2)`

Discussion

148.	Evaluate `lim_(xto0) (e^(sinx)-(1+sinx))/((tan(sinx))^(2))`
Answer» `underset(xto0)lim(e^(sinx)-(1+sinx))/((tan(sinx))^(2))` `=underset(hto0)lim(e^(h)-(1+h))/((tan h)^(2))" "`(where `sin x=h`) `=underset(hto0)lim((1+h+(h^(2))/(2!)+* * * )-(1+h))/((h+(h^(3))/(3)+* * * )^(2))` `=underset(hto0)lim((h^(2))/(2!))/(h^(2))` `=1/2`

Discussion

149.	Evaluate `lim_(xto0) {1^(1//sin^(2)x)+2^(1//sin^(2)x)+...+n^(1//sin^(2)x)}^(sin^(2)x)`.
Answer» Putting `(1)/(sin^(2)x)=t(ge1)`,we have `underset(t tooo)lim(1^(t)+2^(t)+...+n^(t))^(1//t)` `=underset(t tooo)lim(n^(t))^(1//t)[((1)/(n))^(t)+((2)/(n))^(t)+...+1]^(1//t)` `=n underset(t tooo)lim[((1)/(n))^(t)+((2)/(n))^(t)+...+1]^(1//t)` `=n[0+0+...+1]^(0)` `=n`

Discussion

150.	Evaluate `lim_(xto0) (sinx-x)/(x^(3)).`
Answer» `underset(xto0)lim(sinx-x)/(x^(3))" "`(0/0" from") `=underset(xto0)lim((x-(x^(3))/(3!)+x^(5)/(5!)-...)-x)/(x^(3))` `=underset(xto0)lim[-(1)/(3!)+x^(2)/(5!)-... ]=(-1)/(3!)=(-1)/(6)`

Discussion

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