MCQ 2012 Marks
$\lim _{x \rightarrow 0} \frac{e^{\frac{1}{x}}}{e^{\left(\frac{1}{x}+1\right)}}=$
AnswerCorrect option: D. $e ^{-1}$
(D)
$\lim _{x \rightarrow 0} \frac{ e ^{1 / x}}{ e ^{\left(\frac{1}{x}+1\right)}}=\lim _{x \rightarrow 0} \frac{ e ^{1 / x}}{ e ^{\frac{1}{x}} \cdot e }=\lim _{x \rightarrow 0} \frac{1}{ e }= e ^{-1}$
View full question & answer→MCQ 2022 Marks
$\lim _{x \rightarrow 0} \frac{5^x-4^x}{4^x-3^x}$ is equal to
AnswerCorrect option: B. $\frac{\log \left(\frac{5}{4}\right)}{\log \left(\frac{4}{3}\right)}$
(B)
$\lim _{x \rightarrow 0} \frac{5^x-4^x}{4^x-3^x}=\lim _{x \rightarrow 0} \frac{\left(\frac{5^x-1}{x}\right)-\left(\frac{4^x-1}{x}\right)}{\left(\frac{4^x-1}{x}\right)-\left(\frac{3^x-1}{x}\right)}$
$=\frac{\log 5-\log 4}{\log 4-\log 3}$
$=\frac{\log \left(\frac{5}{4}\right)}{\log \left(\frac{4}{3}\right)}$
View full question & answer→MCQ 2032 Marks
$\lim \left(\frac{a^x-b^x}{x}\right)=$
AnswerCorrect option: B. $\log \left(\frac{a}{b}\right)$
(B)
$\lim _{x \rightarrow 0} \frac{ a ^x- b ^x}{x}=\lim _{x \rightarrow 0}\left(\frac{ a ^x-1}{x}\right)-\lim _{x \rightarrow 0}\left(\frac{b^x-1}{x}\right)$
$=\log a -\log b$
$=\log \left(\frac{ a }{ b }\right)$
View full question & answer→MCQ 2042 Marks
The value of $\lim _{x \rightarrow 0} \frac{5^x-5^{-x}}{2 x}=$
Answer(A)
$\lim _{x \rightarrow 0} \frac{5^x-5^{-x}}{2 x}=\lim _{x \rightarrow 0} \frac{5^{2 x}-1}{5^x \cdot 2 x}$
$=\lim _{x \rightarrow 0} \frac{5^{2 x}-1}{2 x} \cdot \frac{1}{5^x}$
$=\log 5\left(\frac{1}{5}\right)^0=\log 5$
View full question & answer→MCQ 2052 Marks
$\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}=$
Answer(B)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{x}$
$=\lim _{x \rightarrow 0}\left(\frac{\cos x}{2 \sqrt{1+\sin x}}+\frac{\cos x}{2 \sqrt{1-\sin x}}\right)$
$=\frac{1}{2}+\frac{1}{2}$
$=1$
Alternate method:
$\lim _{x \rightarrow 0}\left[\frac{1+\sin x-1+\sin x}{x(\sqrt{1+\sin x}+\sqrt{1-\sin x})}\right]$
$=\lim _{x \rightarrow 0} \frac{2 \sin x}{x(\sqrt{1+\sin x}+\sqrt{1-\sin x})}$
$=1$
View full question & answer→MCQ 2062 Marks
$\lim _{x \rightarrow 0} \frac{\tan x}{1-\sqrt{1+\tan x}}=$
Answer(C)
$\lim _{x \rightarrow 0} \frac{\tan x}{1-\sqrt{1+\tan x}}=\lim _{x \rightarrow 0} \frac{\tan x(1+\sqrt{1+\tan x})}{1-1-\tan x}$
$=-\lim _{x \rightarrow 0}(1+\sqrt{1+\tan x})$
$=-(1+\sqrt{1+\tan 0})=-2$
View full question & answer→MCQ 2072 Marks
$\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{\frac{\pi}{2}-\theta}{\cot \theta}=$
Answer(C)
Applying L-Hospital's rule, we get
$\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{\frac{\pi}{2}-\theta}{\cot \theta}=\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{-1}{-\operatorname{cosec}^2 \theta}=1$
View full question & answer→MCQ 2082 Marks
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{2 x-\pi}{\cos x}=$
Answer(C)
$\lim _{x \rightarrow \frac{\pi}{2}} 2\left[\frac{x-\frac{\pi}{2}}{\sin \left(\frac{\pi}{2}-x\right)}\right]=-2$
Alternate method:
Apply L-Hospital's rule.
View full question & answer→MCQ 2092 Marks
$\lim _{x \rightarrow \pi / 6} \frac{\cot ^2 \theta-3}{\operatorname{cosec} \theta-2}=$
Answer(B)
$\lim _{x \rightarrow \pi / 6} \frac{\operatorname{cosec}^2 \theta-4}{\operatorname{cosec} \theta-2}=\lim _{x \rightarrow \pi / 6}(\operatorname{cosec} \theta+2)=4$
View full question & answer→MCQ 2102 Marks
$\lim _{x \rightarrow \alpha} \frac{\sin x-\sin \alpha}{x-\alpha}=$
- A
$0$
- B
- C
$\sin \alpha$
- ✓
$\cos \alpha$
AnswerCorrect option: D. $\cos \alpha$
(D)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow \alpha} \frac{\sin x-\sin \alpha}{x-\alpha}=\lim _{x \rightarrow \alpha} \frac{\cos x}{1}=\cos \alpha$
View full question & answer→MCQ 2112 Marks
$\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}$ is equal to
Answer(B)
$\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}=\lim _{x \rightarrow 0} \frac{2 \sin x(1-\cos x)}{x^3}$
$=2 \lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{(1-\cos x)}{x^2}$
$\lim _{x \rightarrow 0} \frac{1-\cos k x}{x^2}=\frac{ k ^2}{2}$
$=2 \times 1 \times \frac{1^2}{2}$
$=1$
View full question & answer→MCQ 2122 Marks
$\lim _{x \rightarrow 0} \frac{1-\cos x}{\tan ^2 x}=$
- A
- ✓
$\frac{1}{2}$
- C
- D
$\frac{1}{3}$
AnswerCorrect option: B. $\frac{1}{2}$
(B)
$\lim _{x \rightarrow 0} \frac{1-\cos k x}{x^2}=\frac{ k ^2}{2}$
$\lim _{x \rightarrow 0} \frac{1-\cos x}{\tan ^2 x}=\frac{\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}}{\lim _{x \rightarrow 0} \frac{\tan ^2 x}{x^2}}=\frac{\frac{1^2}{2}}{1^2}=\frac{1}{2}$
View full question & answer→MCQ 2132 Marks
$\lim _{x \rightarrow 0} \frac{1-\cos x}{\sin ^2 x}=$
- ✓
$\frac{1}{2}$
- B
$-\frac{1}{2}$
- C
- D
AnswerCorrect option: A. $\frac{1}{2}$
(A)
$\lim _{x \rightarrow 0} \frac{1-\cos k x}{x^2}=\frac{ k ^2}{2}$
$\lim _{x \rightarrow 0} \frac{1-\cos x}{\sin ^2 x}=\frac{\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}}{\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2}}=\frac{\frac{1}{2}}{1^2}=\frac{1}{2}$
Alternate method:
Annly L-Hosnital's rule two times
View full question & answer→MCQ 2142 Marks
$\lim _{x \rightarrow e} \frac{1-\cos x}{x^2}=$
- ✓
$\frac{1}{2}$
- B
- C
$\frac{1}{3}$
- D
AnswerCorrect option: A. $\frac{1}{2}$
(A)
$\lim _{x \rightarrow 0} \frac{1-\cos k x}{x^2}=\frac{ k ^2}{2}$
$\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1^2}{2}=\frac{1}{2}$
View full question & answer→MCQ 2152 Marks
$\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}=$
- A
$0$
- B
- ✓
$\frac{1}{2}$
- D
$\frac{1}{3}$
AnswerCorrect option: C. $\frac{1}{2}$
(C)
$\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}=\lim _{x \rightarrow 0}\left\{\frac{\frac{2 \tan 2 x}{2 x}-1}{3-\frac{\sin x}{x}}\right\}=\frac{1}{2}$
Alternate method:
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x}=\lim _{x \rightarrow 0} \frac{2 \sec ^2 2 x-1}{3-\cos x}$
$=\frac{2-1}{3-1}=\frac{1}{2}$
View full question & answer→MCQ 2162 Marks
$\lim _{x \rightarrow 0} \frac{x^2-\tan 2 x}{\tan x}=$
Answer(B)
$\lim _{x \rightarrow 0} \frac{x^2-\tan 2 x}{\tan x}=\lim _{x \rightarrow 0} \frac{x\left(x-\frac{2 \tan 2 x}{2 x}\right)}{\tan x}=-2$
View full question & answer→MCQ 2172 Marks
$\lim _{\theta \rightarrow 0} \frac{5 \theta \cos \theta-2 \sin \theta}{3 \theta+\tan \theta}=$
- ✓
$\frac{3}{4}$
- B
$-\frac{3}{4}$
- C
$0$
- D
AnswerCorrect option: A. $\frac{3}{4}$
(A)
$\lim _{\theta \rightarrow 0} \frac{5 \cos \theta-\frac{2 \sin \theta}{\theta}}{3+\frac{\tan \theta}{\theta}}=\frac{5-2}{3+1}=\frac{3}{4}$
View full question & answer→MCQ 2182 Marks
$\lim _{x \rightarrow 0} \frac{3 \sin x-\sin 3 x}{x^3}=$
- ✓
- B
- C
$\frac{1}{4}$
- D
$-\frac{1}{4}$
Answer(A)
$\lim _{x \rightarrow 0} \frac{3 \sin x-\sin 3 x}{x^3}=\lim _{x \rightarrow 0} \frac{4 \sin ^3 x}{x^3}=4$
View full question & answer→MCQ 2192 Marks
$\lim _{x \rightarrow 0} \frac{\sin 7 x \cdot \sin 11 x}{5 x^2}=$
- A
$\frac{89}{5}$
- B
$\frac{56}{3}$
- ✓
$\frac{77}{5}$
- D
$\frac{65}{3}$
AnswerCorrect option: C. $\frac{77}{5}$
(C)
$\lim _{x \rightarrow 0} \frac{\sin 7 x \cdot \sin 11 x}{5 x^2}$
$=\frac{1}{5} \lim _{x \rightarrow 0} \frac{\sin 7 x}{x} \times \lim _{x \rightarrow 0} \frac{\sin 11 x}{x}$
$=\frac{1}{5} \lim _{x \rightarrow 0} \frac{\sin 7 x}{7 x} \times 7 \times \lim _{x \rightarrow 0} \frac{\sin 11 x}{11 x} \times 11$
$=\frac{1}{5} \times 7 \times 11$
$=\frac{77}{5}$
View full question & answer→MCQ 2202 Marks
$\lim _{x \rightarrow 0} \frac{\sin \left(x^2+5 x\right)}{x}=$
Answer(D)
$\lim _{x \rightarrow 0} \frac{\sin \left(x^2+5 x\right)}{x}=\lim _{x \rightarrow 0} \frac{\sin \left(x^2+5 x\right)}{x(x+5)} \times(x+5)$
$=1(0+5)$
$=5$
View full question & answer→MCQ 2212 Marks
$\lim _{x \rightarrow 0} \frac{\sin x^{\circ}}{x}=$
- A
- ✓
$\frac{\pi}{180}$
- C
- D
$\frac{\pi}{90}$
AnswerCorrect option: B. $\frac{\pi}{180}$
(B)
$\lim _{x \rightarrow 0} \frac{\sin x^0}{x}=\lim _{x \rightarrow 0} \frac{\sin \frac{\pi x}{180}}{x}=\frac{\pi}{180} \ldots\left[\because x^{\circ}=\frac{\pi x}{180}\right.$ radian $]$
View full question & answer→MCQ 2222 Marks
$\lim _{x \rightarrow 0} \frac{\sin ^3 3 x}{x^3}=$
Answer(B)
$\lim _{x \rightarrow 0} \frac{\sin ^3 3 x}{x^3}=\lim _{x \rightarrow 0} \frac{\sin ^3 3 x}{(3 x)^3} \times 27=27$
View full question & answer→MCQ 2232 Marks
$\lim _{\theta \rightarrow 0} \frac{\sin 3 \theta-\sin \theta}{\sin \theta}=$
- A
- ✓
- C
$\frac{1}{3}$
- D
$\frac{3}{2}$
Answer(B)
$\lim _{v \rightarrow 0} \frac{\sin 3 \theta-\sin \theta}{\sin \theta}=\lim _{v \rightarrow 0} \frac{\sin 3 \theta}{\sin \theta}-\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\sin \theta}$
$=\frac{3}{1}-1=2$
View full question & answer→MCQ 2242 Marks
The value of $\lim _{x \rightarrow 0} \frac{\sin 2 x-\sin 8 x}{x}$ is
Answer(A)
$\lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x} \times 2-\lim _{x \rightarrow 0} \frac{\sin 8 x}{8 x} \times 8=2-8=-6$
View full question & answer→MCQ 2252 Marks
$\lim _{x \rightarrow 0^{+}} \frac{\sin x}{\sqrt{x}}=$
Answer(A)
$\lim _{x \rightarrow 0^{+}} \frac{\sin x}{\sqrt{x}}=\lim _{x \rightarrow 0^{+}}\left(\frac{\sin x}{x}\right) \sqrt{x}$
$=\lim _{x \rightarrow 0^{+}}\left(\frac{\sin x}{x}\right) \lim _{x \rightarrow 0^{+}} \sqrt{x}$
$=1 \times 0$
$=0$
View full question & answer→MCQ 2262 Marks
The value of $\lim _{\theta \rightarrow 0}\left(\frac{\sin \frac{\theta}{5}}{\theta}\right)$ is
AnswerCorrect option: B. $\frac{1}{5}$
(B)
$\lim _{\theta \rightarrow 0} \frac{\sin \frac{\theta}{5}}{\theta}=\lim _{\theta \rightarrow 0} \frac{1}{5} \cdot \frac{\sin \frac{\theta}{5}}{\frac{\theta}{5}}=\frac{1}{5}$
View full question & answer→MCQ 2272 Marks
$\lim _{x \rightarrow 0} \frac{x e^2-\sin x}{x}$ is equal to
Answer(D)
$\lim _{x \rightarrow 0} \frac{x e ^x-\sin x}{x}=\lim _{x \rightarrow 0}\left( e ^x-\frac{\sin x}{x}\right)$
$= e ^0-1$
$=1-1$
$=0$
View full question & answer→MCQ 2282 Marks
$\lim _{x \rightarrow 3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right)=$
Answer(A)
$\lim _{x \rightarrow 3}\left(\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\right)$
$=\lim _{x \rightarrow 3} \frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(\sqrt{x-2}-\sqrt{4-x})(\sqrt{x-2}+\sqrt{4-x})}$
$=\lim _{x \rightarrow 3} \frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2(x-3)}$
$=1$
Alternate method:
Apply L-Hospital's rule.
View full question & answer→MCQ 2292 Marks
$\lim _{x \rightarrow 0}\left(\frac{x}{\sqrt{1+x}-\sqrt{1-x}}\right)$ is equal to
Answer(B)
$\lim _{x \rightarrow 0}\left(\frac{x}{\sqrt{1+x}-\sqrt{1-x}}\right)$
$=\lim _{x \rightarrow 0} \frac{x(\sqrt{1+x}+\sqrt{1-x})}{(1+x)-(1-x)}$
$=\lim _{x \rightarrow 0} \frac{x}{2 x}(\sqrt{1+x}+\sqrt{1-x})$
$=1$
View full question & answer→MCQ 2302 Marks
$\lim _{x \rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2-x}}{x}=$
- A
$\frac{1}{2 \sqrt{2}}$
- ✓
$\frac{1}{\sqrt{2}}$
- C
- D
$\sqrt{2}$
AnswerCorrect option: B. $\frac{1}{\sqrt{2}}$
(B)
$\lim _{x \rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2-x}}{x}$
$=\lim _{x \rightarrow 0} \frac{(\sqrt{2+x}-\sqrt{2-x})(\sqrt{2+x}+\sqrt{2-x})}{x(\sqrt{2+x}+\sqrt{2-x})}$
$=\lim _{x \rightarrow 0} \frac{2}{\sqrt{2+x}+\sqrt{2-x}}$
$=\frac{2}{\sqrt{2+0}+\sqrt{2-0}}=\frac{1}{\sqrt{2}}$
Alternate Method:
$\lim _{x \rightarrow 0} \frac{\sqrt{a+x^n}-\sqrt{a-x^n}}{x^n}=\frac{1}{\sqrt{a}}$
$\lim _{x \rightarrow 0} \frac{\sqrt{2+x}-\sqrt{2-x}}{x}=\frac{1}{\sqrt{2}}$
View full question & answer→MCQ 2312 Marks
$\lim _{x \rightarrow 0} \frac{\sqrt{1+2 x}-1}{x}=$
Answer(D)
By rationalising, we get
$\lim _{x \rightarrow 0} \frac{\sqrt{1+2 x}-1}{x}=\lim _{x \rightarrow 0} \frac{1+2 x-1}{x(\sqrt{1+2 x}+1)}$
$=\lim _{x \rightarrow 0} \frac{2}{\sqrt{1+2 x}+1}$
$=1$
View full question & answer→MCQ 2322 Marks
If the value of $\lim _{x \rightarrow 0} \frac{(1-x)^n-1}{x}$ is 100 , then $n =$
Answer(B)
$\lim _{x \rightarrow 0} \frac{(1-x)^n-1}{x}=100$
$\Rightarrow \lim _{x \rightarrow 0} \frac{(1-x)^n-1^n}{(1-x)-1}=-100$
$\therefore \quad n =-100$
View full question & answer→MCQ 2332 Marks
$\lim _{x \rightarrow 0} \frac{(1+x)^5-1}{(1+x)^3-1}=$
- A
$0$
- B
- ✓
$\frac{5}{3}$
- D
$\frac{5}{3}$
AnswerCorrect option: C. $\frac{5}{3}$
(C)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{(1+x)^5-1}{(1+x)^3-1}=\lim _{x \rightarrow 0} \frac{5(1+x)^4}{3(1+x)^2}=\frac{5}{3}$
View full question & answer→MCQ 2342 Marks
$\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}$ is equal to
AnswerCorrect option: A. $\frac{1}{2}$
(A)
$\lim _{x \rightarrow 0} \frac{(1+x)^{ n }-1}{x}= n$
$\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{x}=\frac{1}{2}$
View full question & answer→MCQ 2352 Marks
$\lim _{x \rightarrow 1} \frac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}}$ is equal to
- A
$\frac{1}{2}$
- B
$\sqrt{2}+1$
- C
- ✓
$1+\frac{1}{\sqrt{2}}$
AnswerCorrect option: D. $1+\frac{1}{\sqrt{2}}$
(D)
$\lim _{x \rightarrow 1} \frac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}}$
$=\lim _{x \rightarrow 1} \frac{\sqrt{(x-1)(x+1)}+\sqrt{x-1}}{\sqrt{(x-1)(x+1)}}$
$=\lim _{x \rightarrow 1} \frac{[\sqrt{x+1}+1]}{\sqrt{x+1}}$
$=\frac{\sqrt{2}+1}{\sqrt{2}}$
$=1+\frac{1}{\sqrt{2}}$
View full question & answer→MCQ 2362 Marks
$\lim _{x \rightarrow 2}\left(\frac{1}{x-2}-\frac{2}{x^2-2 x}\right)=$
- A
$\frac{1}{3}$
- B
$\frac{1}{4}$
- C
$\frac{1}{5}$
- ✓
$\frac{1}{2}$
AnswerCorrect option: D. $\frac{1}{2}$
(D)
$\lim _{x \rightarrow 2}\left(\frac{1}{x-2}-\frac{2}{x^2-2 x}\right)=\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x(x-2)}\right]$
$=\lim _{x \rightarrow 2} \frac{(x-2)}{x(x-2)}$
$=\frac{1}{2}$
View full question & answer→MCQ 2372 Marks
$\lim _{x \rightarrow \sqrt{3}} \frac{x^2-3}{x^2+3 \sqrt{3} x-12}$ is equal to
- A
$\frac{1}{5}$
- ✓
$\frac{2}{5}$
- C
$\frac{\sqrt{3}}{5}$
- D
$\frac{3}{5}$
AnswerCorrect option: B. $\frac{2}{5}$
(B)
$\lim _{x \rightarrow \sqrt{3}} \frac{x^2-3}{x^2+3 \sqrt{3} x-12}=\lim _{x \rightarrow \sqrt{3}} \frac{(x-\sqrt{3})(x+\sqrt{3})}{(x+4 \sqrt{3})(x-\sqrt{3})}$
$=\lim _{x \rightarrow \sqrt{3}} \frac{x+\sqrt{3}}{x+4 \sqrt{3}}$
$=\frac{2}{5}$
View full question & answer→MCQ 2382 Marks
The value of $\lim _{x \rightarrow-2} \frac{\left(x^2-x-6\right)^2}{(x+2)^2}$ is
Answer(B)
$\lim _{x \rightarrow-2} \frac{\left(x^2-x-6\right)^2}{(x+2)^2}=\lim _{x \rightarrow-2} \frac{(x+2)^2(x-3)^2}{(x+2)^2}$
$=(-2-3)^2$
$=25$
View full question & answer→MCQ 2392 Marks
$\lim _{x \rightarrow 2} \frac{x^2-4}{x^2-x-2}=$
- A
$\frac{17}{5}$
- B
$\frac{23}{6}$
- ✓
$\frac{4}{3}$
- D
$\frac{13}{2}$
AnswerCorrect option: C. $\frac{4}{3}$
(C)
Applying L-Hospital's Rule, we get
$\lim _{x \rightarrow 2} \frac{x^2-4}{x^2-x-2}=\lim _{x \rightarrow 2} \frac{2 x}{2 x-1}=\frac{4}{3}$
View full question & answer→MCQ 2402 Marks
$\lim _{x \rightarrow 1} \frac{x-1}{2 x^2-7 x+5}=$
- A
$\frac{1}{3}$
- B
$\frac{1}{11}$
- ✓
$\frac{-1}{3}$
- D
$\frac{-1}{11}$
AnswerCorrect option: C. $\frac{-1}{3}$
(C)
$\lim _{x \rightarrow 1} \frac{x-1}{2 x^2-7 x+5}=\lim _{x \rightarrow 1} \frac{x-1}{(x-1)(2 x-5)}$
$=\lim _{x \rightarrow 1} \frac{1}{2 x-5}=-\frac{1}{3}$
Alternate Method:
Apply L-Hospital's rule.
View full question & answer→MCQ 2412 Marks
$\lim _{x \rightarrow 0} \frac{|x|}{x}=$
Answer(D)
Since $\lim _{x \rightarrow 0^{-}} \frac{|x|}{x}=-1$ and $\lim _{x \rightarrow 0^{+}} \frac{|x|}{x}=1$, limit does not exist.
View full question & answer→MCQ 2422 Marks
$\lim _{x \rightarrow a} \frac{(x+2)^{\frac{5}{3}}-(a+2)^{\frac{5}{3}}}{x-a}=$
- ✓
$\frac{5}{3}(a+2)^{\frac{2}{3}}$
- B
$\frac{5}{3}(a+2)^{\frac{5}{3}}$
- C
$\frac{5}{3} a^{\frac{2}{3}}$
- D
$\frac{5}{3} a^{\frac{5}{3}}$
AnswerCorrect option: A. $\frac{5}{3}(a+2)^{\frac{2}{3}}$
(A)
$\lim _{x \rightarrow a } \frac{(x+2)^{\frac{5}{3}}-( a +2)^{\frac{5}{3}}}{x- a }=\lim _{x \rightarrow a } \frac{(x+2)^{\frac{5}{3}}-( a +2)^{\frac{5}{3}}}{(x+2)-( a +2)}$
$=\frac{5}{3}(a+2)^{\frac{2}{3}}$
View full question & answer→MCQ 2432 Marks
$\lim _{x \rightarrow a} \frac{\sqrt{x}-\sqrt{a}}{x-a}=$
- ✓
$\frac{1}{2 \sqrt{a}}$
- B
$\frac{1}{\sqrt{a}}$
- C
$\frac{2}{\sqrt{a}}$
- D
$\sqrt{a}$
AnswerCorrect option: A. $\frac{1}{2 \sqrt{a}}$
(A)
Applying L-Hospital's Rule, we get
$\lim _{x \rightarrow a } \frac{\sqrt{x}-\sqrt{ a }}{x- a }=\lim _{x \rightarrow a } \frac{1}{2 \sqrt{x}}=\frac{1}{2 \sqrt{ a }}$
View full question & answer→MCQ 2442 Marks
$\lim _{x \rightarrow a} \frac{x^{-1}-a^{-1}}{x-a}=$
- A
$\frac{1}{a}$
- B
$\frac{-1}{ a }$
- C
$\frac{1}{ a ^2}$
- ✓
$\frac{-1}{ a ^2}$
AnswerCorrect option: D. $\frac{-1}{ a ^2}$
(D)
$\lim _{x \rightarrow a } \frac{x^{-1}- a ^{-1}}{x- a }=-1( a )^{-1-1} \quad \ldots\left[\because \lim _{x \rightarrow a } \frac{x^{ n }- a ^{ n }}{x- a }= na ^{ n -1}\right]$
$=\frac{-1}{ a ^2}$
View full question & answer→MCQ 2452 Marks
$\lim _{x \rightarrow 2} \frac{x^2-7 x+2}{x^2+5}=$
- A
$\frac{-2}{5}$
- B
$\frac{-7}{9}$
- ✓
$\frac{-8}{9}$
- D
$\frac{-7}{8}$
AnswerCorrect option: C. $\frac{-8}{9}$
(C)
$\lim _{x \rightarrow 2} \frac{x^2-7 x+2}{x^2+5}=\frac{2^2-7(2)+2}{2^2+5}=\frac{-8}{9}$
View full question & answer→