MCQ 11 Mark
$\lim _{x \rightarrow \infty}\left[\frac{(2 x+3)^7(x-5)^3}{(2 x-5)^{10}}\right]=$
- A
$\frac{3}{8}$
- ✓
$\frac{1}{8}$
- C
$\frac{1}{6}$
- D
$\frac{1}{4}$
AnswerCorrect option: B. $\frac{1}{8}$
(B) $\frac{1}{8}$
Hint:
$\lim _{x \rightarrow \infty} \frac{(2 x+3)^7 \cdot(x-5)^3}{(2 x-5)^{10}} $
$=\frac{\lim _{x \rightarrow \infty}\left(\frac{2 x+3}{x}\right)^7 \cdot\left(\frac{x-5}{x}\right)^3}{\lim _{x \rightarrow \infty}\left(\frac{2 x-5}{x}\right)^{10}} $
$=\frac{\lim _{x \rightarrow \infty}\left(2+\frac{3}{x}\right)^7 \times \lim _{x \rightarrow x}\left(1-\frac{5}{x}\right)^3}{\lim _{x \rightarrow \infty}\left(2-\frac{5}{x}\right)^{10}}$
$=\frac{(2+0)^7 \times(1-0)^3}{(2-0)^{10}} \quad \ldots\left[\lim _{x \rightarrow \infty} \frac{1}{x^k}=0, \mathrm{k}>0\right]$
$=\frac{1}{8}$
View full question & answer→MCQ 21 Mark
$
\lim _{x \rightarrow 3}\left[\frac{5^{x-3}-4^{x-3}}{\sin (x-3)}\right]=
$
- A
$\log 5-4$
- ✓
$\log \frac{5}{4}$
- C
$\frac{\log 5}{\log 4}$
- D
$\frac{\log 5}{4}$
AnswerCorrect option: B. $\log \frac{5}{4}$
(B) $\log \frac{5}{4}$
Hint:
$
\begin{aligned}
& \lim _{x \rightarrow 3} \frac{5^{x-3}-4^{x-3}}{\sin (x-3)} \\
& \text { put } x-3=h \\
\therefore \quad & x=3+h \\
& \text { As } x \rightarrow 3, h \rightarrow 0 \\
\therefore \quad & \text { Required limit } \\
& =\lim _{h \rightarrow 0} \frac{5^h-4^h}{\sinh }\\
& =\lim _{h \rightarrow 0} \frac{\left(5^h-1\right)-\left(4^h-1\right)}{\sin h} \\
& =\frac{\lim _{h \rightarrow 0} \frac{\left(5^h-1\right)}{h}-\frac{\left(4^h-1\right)}{h}}{\lim _{h \rightarrow 0} \frac{\sin h}{h}} \quad \cdots\left[\lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log a\right] \\
& =\frac{\log 5-\log 4}{1} \\
= & \log \left(\frac{5}{4}\right)
\end{aligned}
$
View full question & answer→MCQ 31 Mark
$\lim _{x \rightarrow 0}\left[\frac{\left(3^{\sin x}-1\right)^3}{\left(3^x-1\right) \cdot \tan x \cdot \log (1+x)}\right]=$
- A
$3 \log 3$
- B
$2 \log 3$
- ✓
$(\log 3)^2$
- D
$(\log 3)^3$
AnswerCorrect option: C. $(\log 3)^2$
(C) $(\log 3)^2$
Hint:
$\lim _{x \rightarrow 0} \frac{\left(3^{\operatorname{lin} x}-1\right)^3}{\left(3^x-1\right) \cdot \tan x \cdot \log (1+x)}$
$=\frac{\lim _{x \rightarrow 0} \frac{\left(3^{\sin x}-1\right)^3}{\sin ^3 x} \cdot \frac{\sin ^3 x}{x^3}}{\lim _{x \rightarrow 0}\left(\frac{3^x-1}{x}\right)\left(\frac{\tan x}{x}\right) \cdot \frac{\log (1+x)}{x}} $
$=\frac{\lim _{x \rightarrow 0}\left(\frac{3^{\operatorname{tin} x}-1}{\sin x}\right)^3 \cdot \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^3}{\lim _{x \rightarrow 0}\left(\frac{3^x-1}{x}\right) \cdot \lim \left(\frac{\tan x}{x}\right) \cdot \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}} $
$=(\log 3)^2$
View full question & answer→MCQ 41 Mark
$\lim _{x \rightarrow 0}\left[\frac{x \cdot \log (1+3 x)}{\left(e^{3 x}-1\right)^2}\right]=$
- A
$\frac{1}{e^9}$
- B
$\frac{1}{\mathrm{e}^3}$
- C
$\frac{1}{9}$
- ✓
$\frac{1}{3}$
AnswerCorrect option: D. $\frac{1}{3}$
(D) $\frac{1}{3}$
Hint:
$\lim _{x \rightarrow 0} \frac{x \cdot \log (1+3 x)}{\left(e^{3 x}-1\right)^2}$
$=\frac{\lim _{x \rightarrow 0} \frac{\log (1+3 x)}{x}}{\lim _{x \rightarrow 0}\left(\frac{e^{3 x}-1}{x}\right)^2} $
$=\frac{\lim _{x \rightarrow 0}\left[\frac{\log (1+3 x)}{3 x} \times 3\right]}{\lim _{x \rightarrow 0}\left[\left(\frac{e^{3 x}-1}{3 x}\right)^2 \times(3)^2\right]}=\frac{1}{3}$
View full question & answer→MCQ 51 Mark
$\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{3^{\cos x}-1}{\frac{\pi}{2}-x}\right)=$
- A
$1$
- ✓
$\log 3$
- C
$3^{\frac{\pi}{2}}$
- D
$3 \log 3$
AnswerCorrect option: B. $\log 3$
(B) $\log 3$
Hint:
$ \lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{3^{\cos x}-1}{\frac{\pi}{2}-x}\right) $
$ \text { Put } \frac{\pi}{2}-x=h$
$\therefore \quad x=\frac{\pi}{2}-h$
As $x \rightarrow \frac{\pi}{2}, \mathrm{~h} \rightarrow 0$
Required limit
$=\lim _{h \rightarrow 0} \frac{3^{\cos \left(\frac{\pi}{2}-k\right)}-1}{h}$
$=\lim _{h \rightarrow 0}\left(\frac{3^{\sinh }-1}{\sin h} \times \frac{\sin h}{h}\right)=\log 3$
View full question & answer→MCQ 61 Mark
$\lim _{x \rightarrow 0}\left[\frac{\log (5+x)-\log (5-x)}{\sin x}\right]=$
- A
$\frac{3}{2}$
- B
$-\frac{5}{2}$
- C
$-\frac{1}{2}$
- ✓
$\frac{2}{5}$
AnswerCorrect option: D. $\frac{2}{5}$
(D) $\frac{2}{5}$
Hint:
$\lim _{x \rightarrow 0}\left[\frac{\log (5+x)-\log (5-x)}{\sin x}\right] $
$=\lim _{x \rightarrow 0} \frac{\log \left[5\left(1+\frac{x}{5}\right)\right]-\log \left[5\left(1-\frac{x}{5}\right)\right]}{\sin x} $
$=\lim _{x \rightarrow 0} \frac{\log 5+\log \left(1+\frac{x}{5}\right)-\left[\log 5+\log \left(1-\frac{x}{5}\right)\right]}{\sin x} $
$=\lim _{x \rightarrow 0}\left[\frac{\log \left(1+\frac{x}{5}\right)-\log \left(1-\frac{x}{5}\right)}{x} \times \frac{x}{\sin x}\right] $
$=\lim _{x \rightarrow 0}\left[\frac{\log \left(1+\frac{x}{5}\right)}{5\left(\frac{x}{5}\right)}-\frac{\log \left(1-\frac{x}{5}\right)}{(-5)\left(\frac{-x}{5}\right)}\right] \times \lim _{x \rightarrow 0} \frac{x}{\sin x} $
$=\left[\frac{1}{5}(1)+\frac{1}{5}(1)\right] \times 1=\frac{2}{5}$
View full question & answer→MCQ 71 Mark
$\lim _{x \rightarrow 0}\left(\frac{3+5 x}{3-4 x}\right)^{\frac{1}{x}}=$
- ✓
$\mathrm{e}^3$
- B
$e^6$
- C
$e^9$
- D
$e^{-3}$
AnswerCorrect option: A. $\mathrm{e}^3$
(A) $\mathrm{e}^3$
Hint:
$\lim _{x \rightarrow 0}\left(\frac{3+5 x}{3-4 x}\right)^{\frac{1}{x}} $
$=\lim _{x \rightarrow 0}\left(\frac{1+\frac{5 x}{3}}{1-\frac{4 x}{3}}\right)^{\frac{1}{x}} $$\quad \ldots[\text { Divide numerator and denominator by 3} ]$
$=\frac{\lim _{x \rightarrow 0}\left[\left(1+\frac{5 x}{3}\right)^{\frac{3}{5 x}}\right]^{\frac{5}{3}}}{\lim _{x \rightarrow 0}\left[\left(1-\frac{4 x}{3}\right)^{\frac{-3}{4 x}}\right]^{\frac{-4}{3}}} $
$=\frac{\mathrm{e}^{\frac{5}{3}}}{\mathrm{e}^{\frac{-4}{3}}}=\mathrm{e}^3$
View full question & answer→MCQ 81 Mark
$\lim _{x \rightarrow 0}\left(\frac{15^x-3^x-5^x+1}{\sin ^2 x}\right)=$
- A
$\log 15$
- B
$\log 3+\log 5$
- ✓
$\log 3 . \log 5$
- D
$3 \log 5$
AnswerCorrect option: C. $\log 3 . \log 5$
(C) $\log 3 . \log 5$
Hint:
$\lim _{x \rightarrow 0}\left(\frac{15^x-3^x-5^x+1}{\sin ^2 x}\right)$
$=\lim _{x \rightarrow 0}\left[\frac{\left(5^x-1\right)\left(3^x-1\right)}{x^2} \times \frac{x^2}{\sin ^2 x}\right] $
$\because x \rightarrow 0, x \neq 0$
$\therefore x^2 \neq 0$
$=\log 3 \cdot \log 5$
View full question & answer→MCQ 91 Mark
$\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{3 \cos x+\cos 3 x}{(2 x-\pi)^3}\right]=$
- A
$\frac{3}{2}$
- B
$\frac{1}{2}$
- ✓
$-\frac{1}{2}$
- D
$\frac{1}{4}$
AnswerCorrect option: C. $-\frac{1}{2}$
(C) $-\frac{1}{2}$
Hint:
$\lim _{x-\frac{\pi}{2}} \frac{3 \cos x+\cos 3 x}{(2 x-\pi)^3} $
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{3 \cos x+4 \cos ^3 x-3 \cos x}{(2 x-\pi)^3} $
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{4 \cos ^3 x}{8\left(x-\frac{\pi}{2}\right)^3}$
$Put$ $x=\frac{\pi}{2}+\mathrm{h}$,
$x-\frac{\pi}{2}=\mathrm{h}$
As $x \rightarrow \frac{\pi}{2}, \mathrm{~h} \rightarrow 0$
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{4 \cos ^3 x}{8\left(x-\frac{\pi}{2}\right)^3}$
$=\lim _{h \rightarrow 0} \frac{4 \cos ^3\left(\frac{\pi}{2}+h\right)}{8 h^3}$
$=\lim _{h \rightarrow 0} \frac{4(-\sin h)^3}{8 h^3} \quad \ldots\left[\because \cos \left(\frac{\pi}{2}+\theta\right)=-\sin \theta\right] $
$=-\frac{1}{2} \cdot\left(\lim _{h \rightarrow 0} \frac{\sin h}{h}\right)^3=\frac{-1}{2}$
View full question & answer→MCQ 101 Mark
$\lim _{x \rightarrow 0}\left(\frac{5 \sin x-x \cos x}{2 \tan x-3 x^2}\right)=$
Answer$(C) 2$
Hint:
$\lim _{x \rightarrow 0}\left(\frac{5 \sin x-x \cos x}{2 \tan x-3 x^2}\right) $
$=\frac{\lim _{x \rightarrow 0}\left(\frac{5 \sin x}{x}-\cos x\right)}{\lim _{x \rightarrow 0}\left(\frac{2 \tan x}{x}-3 x\right)} $
$=\frac{5(1)-\cos 0}{2(1)-3(0)}$
$=2$
View full question & answer→MCQ 111 Mark
$\lim _{x \rightarrow \frac{\pi}{3}}\left(\frac{\tan ^2 x-3}{\sec ^3 x-8}\right)=$
- A
$1$
- B
$\frac{1}{2}$
- ✓
$\frac{1}{3}$
- D
$\frac{1}{4}$
AnswerCorrect option: C. $\frac{1}{3}$
(C) $\frac{1}{3}$
Hint:
$\lim _{x \rightarrow \frac{\pi}{3}}\left(\frac{\tan ^2 x-3}{\sec ^3 x-8}\right)$
$=\lim _{x \rightarrow \frac{\pi}{3}} \frac{\sec ^2 x-1-3}{\sec ^3 x-8} $
$=\lim _{x \rightarrow \frac{\pi}{3}} \frac{(\sec x-2)(\sec x+2)}{(\sec x-2)\left(\sec ^2 x+2 \sec x+4\right)} $
$=\lim _{x \rightarrow \frac{\pi}{3}} \frac{\sec x+2}{\sec ^2 x+2 \sec x+4} $
$=\frac{\sec \frac{\pi}{3}+2}{\left(\sec \frac{\pi}{3}\right)^2+2 \sec \frac{\pi}{3}+4} $
$=\frac{2+2}{(2)^2+2(2)+4}=\frac{1}{3} $
View full question & answer→MCQ 121 Mark
$\lim _{x \rightarrow 5}\left(\frac{\sqrt{x+4}-3}{\sqrt{3 x-11-2}}\right)=$
- A
$\frac{-2}{9}$
- B
$\frac{2}{7}$
- C
$\frac{5}{9}$
- ✓
$\frac{2}{9}$
AnswerCorrect option: D. $\frac{2}{9}$
(D) $\frac{2}{9}$
Hint:
$\lim _{x \rightarrow 5}\left(\frac{\sqrt{x+4}-3}{\sqrt{3 x-11}-2}\right)$
$=\lim _{x \rightarrow 5}\left[\frac{\sqrt{x+4}-3}{\sqrt{3 x-11}-2} \times \frac{\sqrt{x+4}+3}{\sqrt{3 x-11}+2} \times \frac{\sqrt{3 x-11}+2}{\sqrt{x+4}+3}\right] $
$=\lim _{x \rightarrow 5} \frac{(x-5)(\sqrt{3 x-11}+2)}{(3 x-15)(\sqrt{x+4}+3)} $
$=\lim _{x \rightarrow 5} \frac{\sqrt{3 x-11}+2}{3(\sqrt{x+4}+3)}$
$=\frac{\sqrt{4}+2}{3(\sqrt{9}+3)}=\frac{2}{9}$
View full question & answer→MCQ 131 Mark
$\lim _{x \rightarrow 3}\left(\frac{1}{x^2-11 x+24}+\frac{1}{x^2-x-6}\right)$
- ✓
$-\frac{2}{25}$
- B
$\frac{2}{25}$
- C
$\frac{7}{25}$
- D
$-\frac{7}{25}$
AnswerCorrect option: A. $-\frac{2}{25}$
(A) $-\frac{2}{25}$
Hint:
$\lim _{x \rightarrow 3}\left(\frac{1}{x^2-11 x+24}+\frac{1}{x^2-x-6}\right)$
$=\lim _{x \rightarrow 3}\left[\frac{1}{(x-8)(x-3)}+\frac{1}{(x-3)(x+2)}\right] $
$=\lim _{x \rightarrow 3} \frac{2 x-6}{(x-8)(x-3)(x+2)}$
$=\lim _{x \rightarrow 3} \frac{2}{(x-8)(x+2)}=\frac{-2}{25}$
View full question & answer→MCQ 141 Mark
$\lim _{x \rightarrow-2}\left(\frac{x^7+128}{x^3+8}\right)=$
- A
$\frac{56}{3}$
- ✓
$\frac{112}{3}$
- C
$\frac{121}{3}$
- D
$\frac{28}{3}$
AnswerCorrect option: B. $\frac{112}{3}$
(B) $\frac{112}{3}$
Hint:
$\lim _{x \rightarrow-2} \frac{x^7+128}{x^3+8} $
$=\frac{\lim _{x \rightarrow-2} \frac{x^7-(-2)^7}{x-(-2)}}{\lim _{x \rightarrow-2} \frac{x^3-(-2)^3}{x-(-2)}}$
$=\frac{7(-2)^6}{3(-2)^2}=\frac{112}{3} \quad \ldots\left[\because \lim _{x \rightarrow \mathrm{a}} \frac{x^n-\mathrm{a}^n}{x-\mathrm{a}}=\mathrm{na}^{\mathrm{n}-1}\right]$
View full question & answer→MCQ 151 Mark
$\lim _{x \rightarrow 2}\left(\frac{x^4-16}{x^2-5 x+6}\right)=$
Answer(C) -32
Hint:
$\lim _{x \rightarrow 2} \frac{x^4-16}{x^2-5 x+6}$
$=\lim _{x \rightarrow 2} \frac{\left(x^2+4\right)(x+2)}{x-3}$
$=\frac{\left[(2)^2+4\right](2+2)}{2-3}=-32$
View full question & answer→MCQ 162 Marks
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$ equals
- A
$\frac{1}{24}$
- ✓
$\frac{1}{16}$
- C
$\frac{1}{8}$
- D
$\frac{1}{4}$
AnswerCorrect option: B. $\frac{1}{16}$
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x(1-\sin x)}{(\pi-2 x)^3}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right)\left[1-\cos \left(\frac{\pi}{2}-x\right)\right]}{(\pi-2 x)^3}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right) 2 \sin ^2\left(\frac{\frac{\pi}{2}-x}{2}\right)}{2\left(\frac{\pi}{2}-x\right)(\pi-2 x)^2}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{2}-x\right)}{2\left(\frac{\pi}{2}-x\right)} \cdot \frac{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{16\left(\frac{\pi}{4}-\frac{x}{2}\right)^2}$
$=\frac{2}{2 \times 16}$
${\left[\because \lim _{x \rightarrow a} \frac{\sin (x-a)}{x-a}=1 \text { and }=\lim _{x \rightarrow a} \frac{\tan (x-a)}{x-a}=1\right]}$
$=\frac{1}{16}$
View full question & answer→MCQ 172 Marks
If $f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$, then as $x$ approaches $a, \frac{g(x) f(a)-g(a) f(x)}{(x-a)}$ approaches
Answer(b) : We have $f(a)=2, f^{\prime}(a)=1, g(a)=-1$ and $g^{\prime}(a)=2$
Let $L=\lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(x)}{x-a}$
It will give $\frac{0}{0}$ form for $x \rightarrow a$,
So, by applying L' Hospital rule, we get
$
\begin{aligned}
L & =\lim _{x \rightarrow a} g^{\prime}(x) f(a)-g(a) f^{\prime}(x) \\
& =g^{\prime}(a) f(a)-g(a) f^{\prime}(a) \\
& =4+1=5
\end{aligned}
$
View full question & answer→MCQ 182 Marks
$\lim _{x \rightarrow \infty} x^3\left\{\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right\}=$
- A
$\frac{1}{\sqrt{2}}$
- ✓
$\frac{1}{4 \sqrt{2}}$
- C
$\frac{-1}{4 \sqrt{2}}$
- D
$\frac{-1}{\sqrt{2}}$
AnswerCorrect option: B. $\frac{1}{4 \sqrt{2}}$
Let $L=\lim _{x \rightarrow \infty} x^3\left\{\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right\}$
$=\lim _{x \rightarrow \infty} x^3\left[x\left\{\sqrt{1+\sqrt{\frac{1}{x^4}+1}}-\sqrt{2}\right\}\right]$
$=\lim _{x \rightarrow \infty} x^4\left\{\sqrt{\left.1+\sqrt{\frac{1}{x^4}+1}-\sqrt{2}\right\}}\right.$
Let $\frac{1}{x^4}=t$, when $x \rightarrow \infty, t \rightarrow 0$
$\therefore L=\lim _{t \rightarrow 0} \frac{\sqrt{1+\sqrt{t+1}}-\sqrt{2}}{t} \quad\left(\frac{0}{0} \text { form }\right)$
Using L'hospital rule, we get
$L=\lim _{t \rightarrow 0} \frac{\frac{1}{2 \sqrt{1+\sqrt{t+1}}} \cdot \frac{1}{2 \sqrt{t+1}}}{1}=\frac{1}{4 \sqrt{2}}$
Hence, the required value is $\frac{1}{4 \sqrt{2}}$.
View full question & answer→MCQ 192 Marks
$\lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}=$
- A
$2 \log 2$
- B
$2(\log 2)^2$
- C
$\frac{1}{2}(\log 2)^2$
- ✓
$(\log 2)^2$
AnswerCorrect option: D. $(\log 2)^2$
(d) : We have, $\lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}$
Given limit is of the form $\frac{0}{0}$. So, applying L' Hopital's rule, we get
$
\lim _{x \rightarrow 1} \frac{2^{2 x-2} \log 2(2)-2^x \log 2+0}{2 \sin (x-1) \cos (x-1)}
$
$
=\lim _{x \rightarrow 1} \frac{2^{2 x-2} 2 \times(\log 2)-2^x \log 2}{\sin 2(x-1)}
$ $\left(\frac{0}{0} \text { form }\right)
$
Again applying L'Hopital's Rule,
$
\lim _{x \rightarrow 1} \frac{2 \log 2 \times 2^{2 x-2}(2 \log 2)-2^x(\log 2)^2}{2 \cos 2(x-1)}
$
$
=\frac{4(\log 2)^2-2(\log 2)^2}{2}=\frac{2(\log 2)^2}{2}=(\log 2)^2
$
View full question & answer→MCQ 202 Marks
$\lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}=$
- A
$\sqrt{5}(\log 3)^2$
- ✓
$8 \sqrt{5}(\log 3)^2$
- C
$16 \sqrt{5} \log 3$
- D
$8 \sqrt{5} \log 3$
AnswerCorrect option: B. $8 \sqrt{5}(\log 3)^2$
$ \lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}$
$=\lim _{x \rightarrow 0} \frac{\left(27^x-9^x-3^x+1\right)(\sqrt{5}+\sqrt{4+\cos x})}{(\sqrt{5})^2-(\sqrt{4+\cos x})^2}$
$=\lim _{x \rightarrow 0} \frac{\left(9^x\left(3^x-1\right)-1\left(3^x-1\right)\right)(\sqrt{5}+\sqrt{4+\cos x})}{1-\cos x}$
$=\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(3^x-1\right)(\sqrt{5}+\sqrt{4+\cos x})}{2 \sin ^2 \frac{x}{2}}$
$=\lim _{x \rightarrow 0} \frac{\frac{\left(9^x-1\right)\left(3^x-1\right)}{x}(\sqrt{5}+\sqrt{4+\cos x})}{\frac{2 \sin ^2 \frac{x}{2}}{4\left(\frac{x}{2}\right)^2}}$
$=\frac{\log 9 \log 3 \cdot 2 \sqrt{5}}{\frac{1}{2}}$
$=8 \sqrt{5}(\log 3)^2$
View full question & answer→MCQ 212 Marks
If $\lim _{x \rightarrow \infty}\left[\frac{x^3+1}{x^2+1}-(a x+b)\right]=2$, then
- A
$a=1$ and $b=1$
- B
$a=1$ and $b=-1$
- ✓
$a=1$ and $b=-2$
- D
$a=1$ and $b=2$
AnswerCorrect option: C. $a=1$ and $b=-2$
(C)
$\lim _{x \rightarrow \infty}\left[\frac{x^3+1}{x^2+1}-(a x+b)\right]=2$
$\Rightarrow \lim _{x \rightarrow \infty}\left(\frac{x^3(1-a)-b x^2-a x+(1-b)}{x^2+1}\right)=2$ ..(i)
Since the limit of the given expression is a finite non-zero number, numerator and denominator are of the same degree.
$\therefore 1-a=0 \Rightarrow a=1$
Putting the value of a in (i), we get
$\lim _{x \rightarrow \infty}\left(\frac{-b x^2-x+1-b}{x^2+1}\right)=2$
$\Rightarrow \lim _{x \rightarrow \infty}\left(\frac{- b -\frac{1}{x}+\frac{1}{x^2}-\frac{ b }{x^2}}{1+\frac{1}{x^2}}\right)=2$
$\Rightarrow- b =2$
$\Rightarrow b =-2$
View full question & answer→MCQ 222 Marks
The value of the constant $\alpha$ and $\beta$ such that $\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-\alpha x-\beta\right)=0$ are respectively
Answer(C)
$\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-\alpha x-\beta\right)=0$
$\Rightarrow \lim _{x \rightarrow \infty} \frac{x^2(1-\alpha)-x(\alpha+\beta)+1-\beta}{x+1}=0$
Since the limit of the given expression is zero, therefore degree of the polynomial in numerator must be less than denominator.
$\therefore \quad 1-\alpha=0$ and $\alpha+\beta=0 \Rightarrow \alpha=1$ and $\beta=-1$.
View full question & answer→MCQ 232 Marks
Let the sequence $< b_n>$ of real numbers satisfy the recurrence relation: $b_{n+1}=\frac{1}{3}\left(2 b_n+\frac{125}{b_n{ }^2}\right)$, $b_n \neq 0$, then $\lim _{n \rightarrow \infty} b_n=$
- A
$0$
- B
$\infty$
- ✓
- D
$\frac{2}{3}$
Answer(C)
Let $\lim _{n \rightarrow \infty} b_n=b$. Tlien,
$b _{ n +1}=\frac{1}{3}\left(2 b_{ n }+\frac{125}{b_{ n }^2}\right)$
$\Rightarrow \lim _{n \rightarrow \infty} b_{n+1}=\frac{1}{3}\left(2 \lim _{n \rightarrow \infty} b_n+\frac{125}{\lim _{n \rightarrow \infty} b_n^2}\right)$
$\Rightarrow b =\frac{1}{3}\left(2 b+\frac{125}{b^2}\right)$
$\Rightarrow b =\frac{125}{b^2}$
$\Rightarrow b^3=125$
$\Rightarrow b =5$
View full question & answer→MCQ 242 Marks
$\lim _{x \rightarrow \infty} \frac{x^n}{ e ^x}=0$ for
- A
no value of $n$
- ✓
- C
$n =0$ only
- D
$n =2$ only
Answer(B)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow \infty} \frac{x^{ n }}{ e ^x}=\lim _{x \rightarrow \infty} \frac{ n x^{ n -1}}{ e ^x}$
$=\lim _{x \rightarrow \infty} \frac{ n ( n -1) x^{ n -2}}{ e ^x}=\lim _{x \rightarrow \infty} \frac{ n !}{ e ^x}=\frac{ n !}{\infty}=0$,
where n is any whole number... $[\because n !$ is defined for all positive integers including zero]
View full question & answer→MCQ 252 Marks
$\lim _{x \rightarrow \infty} \frac{\log x^{ n }-[x]}{[x]}, n \in N ,([x]$ denotes greatest integer less than or equal to $x$ )
Answer(A)
$\lim _{x \rightarrow \infty} \frac{\log x^{ n }-[x]}{[x]}=\lim _{x \rightarrow \infty} \frac{\log x^{ n }}{[x]}-\lim _{x \rightarrow \infty} \frac{[x]}{[x]}$
$=0-1=-1$
View full question & answer→MCQ 262 Marks
$\lim _{x \rightarrow-\infty}\left(\frac{6 x^2-\cos 3 x}{x^2+5}-\frac{5 x^3+3}{\sqrt{x^6+2}}\right)=$
Answer(A)
Put $x=-y$
As $x \rightarrow-\infty, y \rightarrow \infty$
$\lim _{x \rightarrow-\infty}\left(\frac{6 x^2-\cos 3 x}{x^2+5}-\frac{5 x^3+3}{\sqrt{x^6+2}}\right)$
$=\lim _{y \rightarrow \infty} \frac{6 y^2-\cos 3 y}{y^2+5}-\frac{-5 y^3+3}{\sqrt{y^6+2}}$
$=\lim _{y \rightarrow \infty} \frac{6-\frac{\cos 3 y}{y^2}}{1+\frac{5}{y^2}}-\frac{-5+\frac{3}{y^3}}{\sqrt{1+\frac{2}{y^6}}}$
$=\frac{6-0}{1+0}-\frac{-5+0}{\sqrt{1+0}}$
$=11$
View full question & answer→MCQ 272 Marks
The value of $\lim _{x \rightarrow-\infty} \frac{\sqrt{4 x^2+5 x+8}}{4 x+5}$ is
- ✓
$\frac{-1}{2}$
- B
$0$
- C
$\frac{1}{2}$
- D
AnswerCorrect option: A. $\frac{-1}{2}$
(A)
Put $x=-y$
As $x \rightarrow-\infty, y \rightarrow \infty$
$\therefore \lim _{x \rightarrow-\infty} \frac{\sqrt{4 x^2+5 x+8}}{4 x+5}=\lim _{y \rightarrow \infty} \frac{\sqrt{4(-y)^2-5 y+8}}{-4 y+5}$
$=\lim _{y \rightarrow \infty} \frac{\sqrt{4-\frac{5}{y}+\frac{8}{y^2}}}{-4+\frac{5}{y}}$
$=\frac{\sqrt{4}}{-4}=-\frac{1}{2}$
View full question & answer→MCQ 282 Marks
$\lim _{x \rightarrow-\infty} \frac{2 x-1}{\sqrt{x^2+2 x+1}}$ is equal to
Answer(B)
Put $x=-y$
As $x \rightarrow-\infty, y \rightarrow \infty$
$\therefore \quad \lim _{x \rightarrow-\infty} \frac{2 x-1}{\sqrt{x^2+2 x+1}}=\lim _{y \rightarrow \infty} \frac{-2 y-1}{\sqrt{(-y)^2-2 y+1}}$
$=\lim _{y \rightarrow \infty} \frac{-2-\frac{1}{y}}{\sqrt{1-\frac{2}{y}+\frac{1}{y^2}}}=-\frac{2}{1}=-2$
View full question & answer→MCQ 292 Marks
$\lim _{x \rightarrow \infty} \sqrt{\frac{x+\sin x}{x-\cos x}}=$
Answer(B)
$\lim _{x \rightarrow \infty} \sqrt{\left(\frac{x+\sin x}{x-\cos x}\right)}=\lim _{x \rightarrow \infty} \sqrt{\left(\frac{1+\frac{\sin x}{x}}{1-\frac{\cos x}{x}}\right)}$
$=\lim _{x \rightarrow \infty} \sqrt{1}=1$
$\ldots\left[\because \lim _{x \rightarrow \infty} \frac{\sin x}{x}=\lim _{x \rightarrow \infty} \frac{\cos x}{x}=0\right]$
View full question & answer→MCQ 302 Marks
$\lim _{n \rightarrow \infty}\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\ldots+\frac{1}{2^n}\right)$ equals
Answer(C)
$\lim _{n \rightarrow \infty}\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\ldots+\frac{1}{2^n}\right)=\lim _{n \rightarrow \infty} \frac{1}{2} \frac{\left[1-\left(\frac{1}{2}\right)^n\right]}{\left(1-\frac{1}{2}\right)}$
$=\lim _{n \rightarrow \infty}\left(1-\frac{1}{2^n}\right)$
$=1-0=1$
View full question & answer→MCQ 312 Marks
$\lim _{n \rightarrow \infty}\left\{\frac{2}{3}+\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^3+\ldots+\left(\frac{2}{3}\right)^n\right\}$ is equal to
Answer(B)
$\lim _{n \rightarrow \infty}\left[\frac{2}{3}+\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^3+\ldots+\left(\frac{2}{3}\right)^n\right]$
$=\lim _{n \rightarrow \infty} \frac{\frac{2}{3}\left(1-\left(\frac{2}{3}\right)^n\right)}{1-\frac{2}{3}}=\lim _{n \rightarrow \infty} 2\left[1-\left(\frac{2}{3}\right)^n\right]$
$=2(1-0)=2^3$
View full question & answer→MCQ 322 Marks
If $\lim _{n \rightarrow-\infty} \frac{1-(10)^n}{1+(10)^{n+1}}=\frac{-\alpha}{10}$, then the value of $\alpha$ is
Answer(C)
$\lim _{n \rightarrow \infty} \frac{1-(10)^n}{1+(10)^{n+1}}=\lim _{n \rightarrow \infty} \frac{(10)^n\left[\left(\frac{1}{10}\right)^n-1\right]}{(10)^{n+1}\left(1+\frac{1}{10^{n+1}}\right)}$
$=\lim _{n \rightarrow \infty} \frac{\frac{1}{10^n}-1}{10\left(1+\frac{1}{10^{n+1}}\right)}=-\frac{1}{10}$
$\therefore \quad \alpha=1$
View full question & answer→MCQ 332 Marks
$\lim _{n \rightarrow \infty} \frac{3.2^{n-1}-4.5^{n-1}}{5.2^n+7.5^n}=$
- ✓
$\frac{-20}{7}$
- B
$0$
- C
$\frac{3}{5}$
- D
$\frac{-4}{7}$
AnswerCorrect option: A. $\frac{-20}{7}$
(A)
$\lim _{ n \rightarrow \infty} \frac{3.2^{ n +1}-4.5^{ n +1}}{5.2^{ n }+7.5^{ n }}$
$=\lim _{n \rightarrow \infty} \frac{6.2^n-20.5^n}{5.2^n+7.5^n}=\lim _{n \rightarrow \infty} \frac{5^n\left[6(2 / 5)^n-20\right]}{5^n\left[5(2 / 5)^n+7\right]}$
$=-\frac{20}{7} \quad \ldots . .\left[\because n \rightarrow \infty,\left(\frac{2}{5}\right)^n \rightarrow 0\right]$
View full question & answer→MCQ 342 Marks
$\lim _{n \rightarrow \infty}\left(2^n+5^n\right)^{\frac{1}{n}}$ is equal to
AnswerB)
$\lim _{n \rightarrow \infty}\left(2^n+5^n\right)^{\frac{1}{n}}=\lim _{n \rightarrow \infty} 5\left\{1+\left(\frac{2}{5}\right)^n\right\}^{\frac{1}{n}}=5$
Alternate method:
Here, $0<2<5$
If $0 < x < y$, then $\lim _{n \rightarrow \infty}\left(x^n+y^n\right)^{1 / n}=y$
$\therefore \lim _{n \rightarrow \infty}\left(2^n+5^n\right)^{\frac{1}{n}}=5 $
View full question & answer→MCQ 352 Marks
If $0< x< y$, then $\lim _{n \rightarrow \infty}\left(y^n+x^n\right)^{\frac{1}{n}}=$
Answer(C)
Since $0< x< y$
$\therefore \quad 0<\frac{x}{y}<1$
$\therefore \quad \lim _{ n \rightarrow \infty}\left(\frac{x}{y}\right)^{ n }=0$
$\therefore \quad \lim _{n \rightarrow \infty}\left(y^n+x^n\right)^{\frac{1}{n}}=\lim _{n \rightarrow \infty} y\left[1+\left(\frac{x}{y}\right)^n\right]^{\frac{1}{n}}$
$=y(1+0)^0=y$
View full question & answer→MCQ 362 Marks
$\lim _{x \rightarrow \infty} \frac{(x+1)^{10}+(x+2)^{10}+\ldots . .+(x+100)^{10}}{x^{10}+10^{10}}$ is equal to
Answer(D)
$\lim _{x \rightarrow \infty} \frac{(x+1)^{10}+(x+2)^{10}+\ldots \ldots+(x+100)^{10}}{x^{10}+10^{10}}$
$=\lim _{x \rightarrow \infty} \frac{x^{10}\left[\left(1+\frac{1}{x}\right)^{10}+\left(1+\frac{2}{x}\right)^{10}+\ldots+\left(1+\frac{100}{x}\right)^{10}\right]}{x^{10}\left(1+\frac{10^{10}}{x^{10}}\right)}$
$=100$
View full question & answer→MCQ 372 Marks
Let $t_n$ denote the $n^{\text {th }}$ term of the infinite serion $\frac{1}{1!}+\frac{10}{2!}+\frac{21}{3!}+\frac{34}{4!}+\frac{49}{5!}+\ldots$ . Then $\lim _{n \rightarrow \infty} t _{ n }$ is
Answer(B)
Here, $t_n=\frac{n^2+6 n-6}{n!}$
Since denominator is very large compared to numerator.
$\therefore \quad \lim _{n \rightarrow \infty} \frac{n^2+6 n-6}{n!}=0$
View full question & answer→MCQ 382 Marks
The value of
$\lim _{n \rightarrow \infty}\left[\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n-1)(2n+1)}\right]$is equal to
- ✓
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{1}{4}$
- D
AnswerCorrect option: A. $\frac{1}{2}$
(A)
$\lim _{n \rightarrow \infty}\left[\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n-1)(2 n+1)}\right]$
$=\lim _{n \rightarrow \infty} \frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\ldots\right.$. $\left.+\left(\frac{1}{(2 n-1)}-\frac{1}{(2 n+1)}\right)\right]$
$=\lim _{n \rightarrow \infty} \frac{1}{2}\left[1-\frac{1}{2 n+1}\right]$
$=\frac{1}{2}$
View full question & answer→MCQ 392 Marks
$\lim _{n \rightarrow \infty} \frac{1^4+2^4+3^4+\ldots+n^4}{n^5} -\lim _{n \rightarrow \infty} \frac{1^3+2^3+3^3+\ldots+n^3}{n^3} \text { is }$
- ✓
$\frac{1}{5}$
- B
$\frac{1}{30}$
- C
$0$
- D
$\frac{1}{4}$
AnswerCorrect option: A. $\frac{1}{5}$
(A)
$\lim _{n \rightarrow \infty} \frac{1^4+2^4+\ldots .+n^4}{n^5}-\lim _{n \rightarrow \infty} \frac{1^3+2^3+\ldots .+n^3}{n^5}$
$\lim _{x \rightarrow \infty} \frac{1^\alpha+2^\alpha+3^\alpha+\ldots+x^\alpha}{x^{\alpha+1}}=\frac{1}{\alpha+1}$
$=\frac{1}{4+1}-\lim _{n \rightarrow \infty} \frac{n^2(n+1)^2}{4 n^5} $
$=\frac{1}{5}-\frac{1}{4} \lim _{n \rightarrow \infty} \frac{1}{n}\left(1+\frac{1}{n}\right)^2$
$=\frac{1}{5}-0=\frac{1}{5}$
View full question & answer→MCQ 402 Marks
$\lim _{n \rightarrow \infty}\left[\frac{1}{n^3+1}+\frac{4}{n^3+1}+\frac{9}{n^3+1}+\ldots+\frac{n^2}{n^3+1}\right]=$
- A
- B
$\frac{2}{3}$
- ✓
$\frac{1}{3}$
- D
$0$
AnswerCorrect option: C. $\frac{1}{3}$
(C)
Given limit $=\lim _{n \rightarrow \infty}\frac{1^2+2^2+3^2+\ldots+n^2}{1+n^3}$
$=\lim _{n \rightarrow \infty} \frac{\sum n^2}{1+n^3}$
$=\lim _{n \rightarrow \infty} \frac{1}{6} \frac{n(n+1)(2 n+1)}{1+n^3}$
$=\lim _{n \rightarrow \infty} \frac{1}{6} \frac{\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)}{\left(\frac{1}{n^3}+1\right)}$
$=\frac{1}{6} \cdot 1 \cdot \frac{2}{(1)}$
$=\frac{1}{3}$
View full question & answer→MCQ 412 Marks
The value of $\lim _{n \rightarrow\infty} \frac{1.2+2.3+3.4+\ldots+n(n+1)}{n^3}$ is
AnswerCorrect option: C. $\frac{1}{3}$
(C)
$\lim _{n \rightarrow \infty} \frac{1.2+2.3+3.4+\ldots+n(n+1)}{n^3}$
$=\lim _{n \rightarrow \infty} \frac{\sum_{r=1}^n r(r+1)}{n^3}$
$=\lim _{n \rightarrow \infty} \frac{\sum_{r=1}^n r^2+\sum_{r=1}^n r}{n^3}$
$=\lim _{n \rightarrow \infty} \frac{\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}}{n^3}$
$=\lim _{n \rightarrow \infty} \frac{n(n+1)(n+2)}{3 n^3}=\frac{1}{3}$
View full question & answer→MCQ 422 Marks
$\lim _{x \rightarrow \infty} \frac{1+2+3+\ldots+n}{1+3+5+\ldots+(2 n-1)}=$
- A
- B
$\frac{3}{2}$
- ✓
$\frac{1}{2}$
- D
AnswerCorrect option: C. $\frac{1}{2}$
(C)
$\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{1+3+5+\ldots+(2 n-1)}=\lim _{n \rightarrow \infty} \frac{\frac{n(n+1)}{2}}{n^2}$
$=\lim _{n \rightarrow \infty} \frac{n(n+1)}{2 n^2}$
$=\lim _{n \rightarrow \infty} \frac{n^2+n}{2 n^2}=\frac{1}{2}$
View full question & answer→MCQ 432 Marks
$\sum_{r=1}^n(2 r-1)=x, \text { then }$
$\lim _{n \rightarrow \infty}\left[\frac{1^3}{x^2}+\frac{2^3}{x^2}+\frac{3^3}{x^2}+\ldots+\frac{n^3}{x^2}\right]=$
- ✓
$\frac{1}{4}$
- B
- C
$\frac{1}{2}$
- D
AnswerCorrect option: A. $\frac{1}{4}$
(A)
$\sum_{ r =1}^{ n }(2 r -1)=x$
$\therefore x=1+3+5+\ldots+(2 n-1)=n^2$
$\Rightarrow x^2= n ^4$
$\therefore \lim _{n \rightarrow \infty}\left[\frac{1^3+2^3+\ldots+n^3}{x^2}\right]$
$=\lim _{n \rightarrow \infty}\left[\frac{1^3+2^3+\ldots+n^3}{n^4}\right]$
$\lim _{x \rightarrow \infty} \frac{1^\alpha+2^\alpha+3^\alpha+\ldots+x^\alpha}{x^{\alpha+1}}=\frac{1}{\alpha+1}$
$=\frac{1}{3+1}=\frac{1}{4}$
View full question & answer→MCQ 442 Marks
The value of $\lim _{x \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2+100}$ is equal to
- A
$\infty$
- ✓
$\frac{1}{2}$
- C
- D
$0$
AnswerCorrect option: B. $\frac{1}{2}$
(B)
$\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots \ldots+n}{n^2+100}$
$=\lim _{n \rightarrow \infty} \frac{n(n+1)}{2\left(n^2+100\right)}$
$=\lim _{n \rightarrow \infty} \frac{n^2\left(1+\frac{1}{n}\right)}{2 n^2\left(1+\frac{100}{n^2}\right)}=\frac{1}{2}$
View full question & answer→MCQ 452 Marks
$\lim _{n \rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\ldots+\frac{n}{n^2}\right)$ is
- ✓
$\frac{1}{2}$
- B
- C
- D
$-\frac{1}{2}$
AnswerCorrect option: A. $\frac{1}{2}$
(A)
$\lim _{n \rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\ldots+\frac{n}{n^2}\right)$
$=\lim _{n \rightarrow \infty}\left(\frac{1+2+3+\ldots+n}{n^2}\right)=\lim _{n \rightarrow \infty} \frac{\frac{n(n+1)}{2}}{n^2}$
$=\frac{1}{2} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=\frac{1}{2}$
Alternate method:
$\lim _{n \rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\ldots .+\frac{n}{n^2}\right)$
$=\lim _{n \rightarrow \infty}\left(\frac{1+2+3+\ldots+n}{n^2}\right)$
$\lim _{x \rightarrow \infty} \frac{1^\alpha+2^\alpha+3^\alpha+\ldots+x^\alpha}{x^{\alpha+1}}=\frac{1}{\alpha+1}$
$=\frac{1}{1+1} $
$=\frac{1}{2}$
View full question & answer→MCQ 462 Marks
The value of $\lim _{n \rightarrow \infty} \frac{1-n^2}{\sum n}$ will be
Answer(A)
$\lim _{n \rightarrow \infty} \frac{1-n^2}{\sum n}=\lim _{n \rightarrow \infty} \frac{(1-n)(1+n)}{\frac{n(n+1)}{2}}=\lim _{n \rightarrow \infty} \frac{2(1-n)}{n}$
$=2 \lim _{n \rightarrow \infty}\left(\frac{1}{n}-1\right)=2(-1)=-2$
View full question & answer→MCQ 472 Marks
If $x_{ n }=\frac{(1-2+3-4+5-6+\ldots)-2 n}{\sqrt{ n ^2+1}+\sqrt{4 n ^2-1}}$, then $\lim _{ n \rightarrow \infty} x_{ n }$ is equal to
- A
$\frac{1}{3}$
- ✓
$-\frac{2}{3}$
- C
$\frac{2}{3}$
- D
AnswerCorrect option: B. $-\frac{2}{3}$
(B)
$\lim _{n \rightarrow \infty} x_n=\lim _{n \rightarrow \infty} \frac{(1-2+3-4+5-6+\ldots)-2 n}{\sqrt{n^2+1}+\sqrt{4 n^2-1}}$
$=\lim _{n \rightarrow \infty} \frac{n\left[\left(\frac{1}{n}-\frac{2}{n}+\frac{3}{n}-\ldots\right)-2\right]}{n\left[\sqrt{1+\frac{1}{n^2}}+\sqrt{4-\frac{1}{n^2}}\right]}=\frac{-2}{1+\sqrt{4}}=\frac{-2}{3}$
View full question & answer→MCQ 482 Marks
$\lim _{x \rightarrow\infty} \frac{\sqrt{x^2+a^2}-\sqrt{x^2+b^2}}{\sqrt{x^2+c^2}-\sqrt{x^2+d^2}}=$
- ✓
$\frac{a^2-b^2}{c^2-d^2}$
- B
$\frac{a^2+b^2}{c^2-d^2}$
- C
$\frac{a^2+b^2}{c^2+d^2}$
- D
$\frac{a^2-b^2}{c^2+d^2}$
AnswerCorrect option: A. $\frac{a^2-b^2}{c^2-d^2}$
(A)
$\lim _{x \rightarrow \infty} \frac{\sqrt{x^2+ a ^2}-\sqrt{x^2+ b ^2}}{\sqrt{x^2+ c ^2}-\sqrt{x^2+ d ^2}}$
$=\lim _{x \rightarrow \infty} \frac{\left( a ^2- b ^2\right)}{\left( c ^2- d ^2\right)}\left[\frac{\sqrt{1+\frac{ c ^2}{x^2}}+\sqrt{1+\frac{ d ^2}{x^2}}}{\sqrt{1+\frac{ a ^2}{x^2}+\sqrt{1+\frac{ b ^2}{x^2}}}}\right]$
$=\frac{ a ^2- b ^2}{ c ^2- d ^2}$
View full question & answer→MCQ 492 Marks
$\lim _{x \rightarrow \infty}(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x})$ is equal to
- A
$0$
- ✓
$\frac{1}{2}$
- C
$\log 2$
- D
$e^4$
AnswerCorrect option: B. $\frac{1}{2}$
(B)
$\lim _{x \rightarrow \infty}(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x})$
$=\lim _{x \rightarrow \infty} \frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}$
$=\lim _{x \rightarrow \infty} \frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}$
$=\lim _{x \rightarrow \infty} \frac{\sqrt{1+x^{-1 / 2}}}{\sqrt{1+\sqrt{x^{-1}+x^{-3 / 2}}}+1}=\frac{1}{2}$
View full question & answer→MCQ 502 Marks
$\lim _{x \rightarrow \infty}\left(\sqrt{x^2+8 x+3}-\sqrt{x^2+4 x+3}\right)=$
- A
$0$
- B
$\infty$
- ✓
- D
$\frac{1}{2}$
Answer(C)
On rationalising, we get
$\lim _{x \rightarrow \infty}\left(\sqrt{x^2+8 x+3}-\sqrt{x^2+4 x+3}\right)$
$=\lim _{x \rightarrow \infty} \frac{4 x}{\sqrt{x^2+8 x+3}+\sqrt{x^2+4 x+3}}$
$=\lim _{x \rightarrow \infty} \frac{4}{\left(\sqrt{1+\frac{8}{x}+\frac{3}{x^2}}+\sqrt{1+\frac{4}{x}+\frac{3}{x^2}}\right)}=2$
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