MCQ 1012 Marks
$\lim _{x \rightarrow \frac{\pi}{4}} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}=$
- ✓
$5 \sqrt{2}$
- B
$3 \sqrt{2}$
- C
$\sqrt{2}$
- D
$\frac{5 \sqrt{2}}{2}$
AnswerCorrect option: A. $5 \sqrt{2}$
(A)
$\lim _{x \rightarrow \pi / 4} \frac{4 \sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2 x}$
$=\lim _{x \rightarrow \pi / 4} \frac{\left[(\cos x+\sin x)^2\right]^{\frac{5}{2}}-(2)^{\frac{5}{2}}}{(1+\sin 2 x) 2}$
$=\lim _{x \rightarrow \pi / 4} \frac{(1+\sin 2 x)^{\frac{5}{2}}-2^{\frac{5}{2}}}{(1+\sin 2 x)-2}$
$=\lim _{y \rightarrow 2} \frac{y^{\frac{5}{2}}-2^{\frac{5}{2}}}{y-2}$, where $y=1+\sin 2 x$
$=\frac{5}{2} \times 2^{\frac{5}{2}-1}=5 \sqrt{2}$
View full question & answer→MCQ 1022 Marks
$\lim _{x \rightarrow \frac{\pi}{3}} \frac{3 \tan x-\tan ^3 x}{\cos \left(x+\frac{\pi}{6}\right)}$ is equal to
Answer(C)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow \frac{\pi}{3}} \frac{3 \tan x-\tan ^3 x}{\cos \left(x+\frac{\pi}{6}\right)}$
$=\lim _{x \rightarrow \frac{\pi}{3}} \frac{3 \sec ^2 x-3 \tan ^2 x \cdot \sec ^2 x}{-\sin \left(x+\frac{\pi}{6}\right)}$
$=\frac{3 \sec ^2 \frac{\pi}{3}-3 \tan ^2 \frac{\pi}{3} \sec ^2 \frac{\pi}{3}}{-\sin \frac{\pi}{2}}=24$
View full question & answer→MCQ 1032 Marks
$\lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}$ equals
- A
$\sqrt{3}$
- ✓
$\frac{1}{\sqrt{3}}$
- C
$-\sqrt{3}$
- D
$-\frac{1}{\sqrt{3}}$
AnswerCorrect option: B. $\frac{1}{\sqrt{3}}$
(B)
Applying L-Hospital's Rule, we get
$\lim _{x \rightarrow \frac{\pi}{6}} \frac{3 \sin x-\sqrt{3} \cos x}{6 x-\pi}=\lim _{x \rightarrow \frac{\pi}{6}}\left(\frac{3 \cos x+\sqrt{3} \sin x}{6}\right)$
$=\frac{3 \cdot \frac{\sqrt{3}}{2}+\sqrt{3 \cdot \frac{1}{2}}}{6}$
$=\frac{1}{\sqrt{3}}$
View full question & answer→MCQ 1042 Marks
$\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}=$
Answer(B)
$\lim _{x \rightarrow \pi / 6} \frac{2 \sin ^2 x+\sin x-1}{2 \sin ^2 x-3 \sin x+1}$
$=\lim _{x \rightarrow \pi / 6} \frac{(2 \sin x-1)(\sin x+1)}{(2 \sin x-1)(\sin x-1)}$
$=\lim _{x \rightarrow \pi/6} \frac{\sin x+1}{\sin x-1}=-3$
View full question & answer→MCQ 1052 Marks
$\lim _{x \rightarrow \pi} \frac{1+\cos ^3 x}{(x-\pi)^2}$ is equal to
- A
$\frac{1}{3}$
- B
$\frac{1}{2}$
- ✓
$\frac{3}{2}$
- D
AnswerCorrect option: C. $\frac{3}{2}$
(C)
$\lim _{x \rightarrow \pi} \frac{1+\cos ^3 x}{(x-\pi)^2}$
$=\lim _{h \rightarrow 0} \frac{1+\cos ^3(\pi+h)}{h^2}$
$=\lim _{h \rightarrow 0} \frac{1-\cos ^3 h}{h^2}$
$=\lim _{h \rightarrow 0}\left(\frac{1-\cos h}{h^2}\right) \cdot \lim _{h \rightarrow 0}\left(1+\cos h+\cos ^2 h\right)$
$=\frac{1}{2}(1+1+1)=\frac{3}{2}$
View full question & answer→MCQ 1062 Marks
$\lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{(\pi-2 x)^2}=$
AnswerCorrect option: D. $\frac{1}{2}$
(D)
Put $\pi-2 x=\theta$
$\Rightarrow 2 x=\pi-\theta$ and as $x \rightarrow \frac{\pi}{2}, \theta \rightarrow 0$
$\therefore \quad \lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{(\pi-2 x)^2}=\lim _{\theta \rightarrow 0} \frac{1+\cos (\pi-\theta)}{\theta^2}$
$=\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^2}=\lim _{\theta \rightarrow 0} \frac{2 \sin ^2 \frac{\theta}{2}}{\frac{\theta^2}{4} \times 4}=\frac{1}{2} \lim _{\theta \rightarrow 0}\left(\frac{\sin \frac{\theta}{2}}{\frac{\theta}{2}}\right)^2$
$=\frac{1}{2}( l )^2=\frac{1}{2}$
View full question & answer→MCQ 1072 Marks
$\lim _{x \rightarrow 1} \frac{1+\cos \pi x}{\tan ^2 \pi x}$ is equal to
AnswerCorrect option: B. $\frac{1}{2}$
(B)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 1} \frac{1+\cos \pi x}{\tan ^2 \pi x}=\lim _{x \rightarrow 1} \frac{-\pi \sin \pi x}{2 \pi \tan \pi x \sec ^2 \pi x}$
$=\frac{-1}{2} \lim _{x \rightarrow 1} \cos ^3 \pi x$
$=-\frac{1}{2}(-1)^3=\frac{1}{2}$
View full question & answer→MCQ 1082 Marks
$\lim _{\alpha \rightarrow \frac{\pi}{4}} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}}=$
- ✓
$\sqrt{2}$
- B
$\frac{1}{\sqrt{2}}$
- C
- D
AnswerCorrect option: A. $\sqrt{2}$
(A)
$\lim _{\alpha \rightarrow \frac{\pi}{4}} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}}$
$=\lim _{\alpha \rightarrow \frac{\pi}{4}}\left\{\frac{\sqrt{2}\left(\sin \alpha \cdot \frac{1}{\sqrt{2}}-\cos \alpha \cdot \frac{1}{\sqrt{2}}\right)}{\left(\alpha-\frac{\pi}{4}\right)}\right\}$
$=\sqrt{2} \lim _{\alpha \rightarrow \frac{\pi}{4}} \frac{\sin \left(\alpha-\frac{\pi}{4}\right)}{\left(\alpha-\frac{\pi}{4}\right)}=\sqrt{2}(1)=\sqrt{2}$
View full question & answer→MCQ 1092 Marks
$\lim _{\alpha \rightarrow \beta} \frac{\sin ^2 \alpha-\sin ^2 \beta}{\alpha^2-\beta^2}=$
AnswerCorrect option: D. $\frac{\sin 2 \beta}{2 \beta}$
(D)
Applying L-Hospital's rule, we get
$\lim _{\alpha \rightarrow \beta} \frac{\sin ^2 \alpha-\sin ^2 \beta}{\alpha^2-\beta^2}=\lim _{\alpha \rightarrow \beta} \frac{2 \sin \alpha \cos \alpha}{2 \alpha}$
$=\lim _{\alpha \rightarrow \beta} \frac{\sin 2 \alpha}{2 \alpha}$
$=\frac{\sin 2 \beta}{2 \beta}$
View full question & answer→MCQ 1102 Marks
$\lim _{x \rightarrow \pi / 4} \frac{\sqrt{2} \cos x-1}{\cot x-1}=$
- A
$\frac{1}{\sqrt{2}}$
- ✓
$\frac{1}{2}$
- C
$\frac{1}{2 \sqrt{2}}$
- D
AnswerCorrect option: B. $\frac{1}{2}$
(B)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2} \sin x}{-\operatorname{cosec}^2 x}$
$=\frac{\sqrt{2} \times \frac{1}{\sqrt{2}}}{(\sqrt{2})^2}=\frac{1}{2}$
View full question & answer→MCQ 1112 Marks
$\lim _{x \rightarrow a} \frac{\cos x-\cos a}{\cot x-\cot a}=$
AnswerCorrect option: C. $\sin ^3 a$
(C)
Applying L Hospital’s Rule, we get
$\lim _{x \rightarrow a} \frac{\cos x-\cos a}{\cot x-\cot a}=\lim _{x \rightarrow a}\left(\frac{-\sin x}{-\operatorname{cosec}^2 x}\right)$
$=\lim _{x \rightarrow a } \sin ^3 x=\sin ^3 a$
View full question & answer→MCQ 1122 Marks
$\lim _{\theta \rightarrow \frac{\pi}{2}}(\sec \theta-\tan \theta)=$
- ✓
$0$
- B
$\frac{1}{2}$
- C
- D
$\infty$
Answer(A)
$\lim _{\theta \rightarrow \frac{\pi}{2}}(\sec \theta-\tan \theta)=\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{1-\sin \theta}{\cos \theta}$
$=\lim _{\theta \rightarrow \frac{\pi}{2}} \frac{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)^2}{\left(\cos \frac{\theta}{2}-\sin \frac{\theta}{2}\right)\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)}=0$
View full question & answer→MCQ 1132 Marks
$\lim _{x \rightarrow \frac{\pi}{2}} \tan x \log \sin x=$
Answer(A)
$\lim _{x \rightarrow \frac{\pi}{2}} \tan x \log \sin x=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\log \sin x}{\cot x}$
$=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\frac{1}{\sin x} \cos x}{-\operatorname{cosec}^2 x}=0$
...[Applying L-Hospital's rule]
View full question & answer→MCQ 1142 Marks
If $\lim _{x \rightarrow 0} \frac{\cos 4 x+ a \cos 2 x+ b }{x^4}$ is finite, then $( a , b )$ =
Answer(C)
Since $\lim _{x \rightarrow 0} \frac{\cos 4 x+ a \cos 2 x+ b }{x^4}$ is finite.
$\therefore \quad \frac{\cos 4 x+ a \cos 2 x+ b }{x^4}$ should be of the form $\frac{0}{0}$ at $x=0$.
Consequently, the value $\cos 4 x+ a \cos 2 x+ b$ must be zero at $x=0$
i.e., $1+a+b=0$ ...(i)
Applying L-Hospital's rule in the given limit, we get
$\lim _{x \rightarrow 0} \frac{-4 \sin 4 x-2 a \sin 2 x}{4 x^3}$
$=\lim _{x \rightarrow 0} \frac{-16 \cos 4 x-4 a \cos 2 x}{12 x^2}$
This should be of the form $\frac{0}{0}$.
$\therefore \quad-16-4 a=0 \Rightarrow a=-4$
Putting the value of a in (i), we get
$b=3$
View full question & answer→MCQ 1152 Marks
If $\lim _{x \rightarrow 0} \frac{2 \operatorname{asin} x-\sin 2 x}{\tan ^3 x}$ exists and is equal to 1 then the value of $a$ is
Answer(B)
$\lim _{x \rightarrow 0} \frac{2 a \sin x-\sin 2 x}{\tan ^3 x}=1$
$\Rightarrow \lim _{x \rightarrow 0} \frac{2 a \left(x-\frac{x^3}{3!}+\ldots .\right)-\left(2 x-\frac{8 x^3}{3!}+\ldots .\right)}{x^3+\ldots .}=1$
$\Rightarrow \lim _{x \rightarrow 0} \frac{2( a -1) x+\left(\frac{4}{3}-\frac{ a }{3}\right) x^3+\ldots}{x^3+\ldots}=1$
$\Rightarrow a -1=0 \Rightarrow a =1$
View full question & answer→MCQ 1162 Marks
If $\lim _{x \rightarrow 0} \frac{\{(a-n) n x-\tan x\} \sin n x}{x^2}=0$, where $n$ is a non-zero real number, then $a$ is equal to
- A
$0$
- B
$\frac{n+1}{n}$
- C
- ✓
$n+\frac{1}{n}$
AnswerCorrect option: D. $n+\frac{1}{n}$
(D)
$\lim _{x \rightarrow 0} \frac{\{( a - n ) n x-\tan x\} \sin n x}{x^2}-0$
$\Rightarrow \lim _{x \rightarrow 0}\left\{\frac{( a - n ) n x}{x}-\frac{\tan x}{x}\right\} \frac{\sin n x}{x}=0$
$\Rightarrow \lim _{x \rightarrow 0}\left\{( a - n ) n -\frac{\tan x}{x}\right\} \frac{\sin n x}{ n x} \times n =0$
$\Rightarrow\lfloor(a-n) n-1\rfloor n=0$
$\Rightarrow(a-n) n=1 \Rightarrow a=n+\frac{1}{n}$
View full question & answer→MCQ 1172 Marks
$\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2}$ is
- A
- B
$-2$
- ✓
$\frac{1}{2}$
- D
$-\frac{1}{2}$
AnswerCorrect option: C. $\frac{1}{2}$
(C)
$\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^2}$
$=\lim _{x \rightarrow 0} \frac{x(\tan 2 x-2 \tan x)}{\left(2 \sin ^2 x\right)^2}$
$=\lim _{x \rightarrow 0} \frac{x(\tan 2 x-2 \tan x)}{4 \sin ^4 x}$
$=\frac{1}{4} \lim _{x \rightarrow 0} \frac{x\left\{\left(2 x+\frac{1}{3}(2 x)^3+\frac{2}{15}(2 x)^5+\ldots\right)-2\left(x+\frac{x^3}{3}+\frac{2}{15} x^5+\ldots\right)\right\}}{x^4\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\ldots\right)^4}$
$=\frac{1}{4}\left(\frac{8}{3}-\frac{2}{3}\right)=\frac{2}{4}=\frac{1}{2}$
View full question & answer→MCQ 1182 Marks
$\lim _{x \rightarrow 0} \frac{\sin x-x}{x^3}=$
- A
$\frac{1}{3}$
- B
$-\frac{1}{3}$
- C
$\frac{1}{6}$
- ✓
$-\frac{1}{6}$
AnswerCorrect option: D. $-\frac{1}{6}$
(D)
$\lim _{x \rightarrow 0} \frac{\sin x-x}{x^3}$
$=\lim _{x \rightarrow 0} \frac{-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots}{x^3}$
$\ldots\left[\because \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots.\right]$
$=\lim _{x \rightarrow 0}\left(-\frac{1}{3!}+\frac{x^2}{5!}-\ldots\right)=\frac{-1}{3!}=\frac{-1}{6}$
View full question & answer→MCQ 1192 Marks
$\lim _{x \rightarrow 0} \frac{\sin x-x+\frac{x^3}{6}}{x^5}=$
- ✓
$\frac{1}{120}$
- B
$-\frac{1}{120}$
- C
$\frac{1}{20}$
- D
AnswerCorrect option: A. $\frac{1}{120}$
(A)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{\sin x-x+\frac{x^3}{6}}{x^5}=\lim _{x \rightarrow 0} \frac{\cos x-1+\frac{3 x^2}{6}}{5 x^4}$
$=\lim _{x \rightarrow 0} \frac{-\sin x+\frac{6 x}{6}}{20 x^3}=\lim _{x \rightarrow 0} \frac{-\cos x+1}{60 x^2}=\lim _{x \rightarrow 0} \frac{\sin x}{120 x}$
$=\lim _{x \rightarrow 0} \frac{\cos x}{120}=\frac{1}{120}$
View full question & answer→MCQ 1202 Marks
The value of $\lim _{x \rightarrow 2} \frac{\sqrt{1-\cos 2(x-2)}}{x-2}$ is
- A
$\sqrt{2}$
- B
$-\sqrt{2}$
- C
- ✓
Answer(D)
$\lim _{x \rightarrow 2} \frac{\sqrt{1-\cos 2(x-2)}}{x-2}=\lim _{x \rightarrow 2} \frac{|\sqrt{2} \sin (x-2)|}{x-2}$
Now,
$\lim _{x \rightarrow 2^{+}} \frac{\sqrt{1-\cos 2(x-2)}}{x-2}=\lim _{x \rightarrow 2^{+}} \frac{|\sqrt{2} \sin (x-2)|}{x-2}$
$=\sqrt{2} \lim _{x \rightarrow 2} \frac{\sin (x-2)}{x-2}$
$\ldots\left[\begin{array}{c}\because x>2 \Rightarrow x-2>0 \Rightarrow \sin (x-2)>0 \\ \Rightarrow|\sin (x-2)|=\sin (x-2)\end{array}\right]$
$=\sqrt{2}(1)=\sqrt{2}$ and
$\lim _{x \rightarrow 2^{-}} \frac{\sqrt{1-\cos 2(x-2)}}{x-2}=\lim _{x \rightarrow 2^{-}} \frac{|\sqrt{2} \sin (x-2)|}{x-2}$
$=-\sqrt{2} \lim _{x \rightarrow 2} \frac{\sin (x-2)}{x-2}$
$\ldots\left[\begin{array}{r}\because x<2 \Rightarrow x-2<0 \Rightarrow \sin (x-2)<0 \\ \Rightarrow|\sin (x-2)|=-\sin (x-2)\end{array}\right]$
$=-\sqrt{2}(1)=-\sqrt{2}$
Hence, $\lim _{x \rightarrow 2} \frac{\sqrt{1-\cos 2(x-2)}}{x-2}$ does not exist.
View full question & answer→MCQ 1212 Marks
$\lim _{x \rightarrow 0} \frac{\sqrt{\frac{1}{2}(1-\cos 2 x)}}{x}=$
Answer(D)
$\lim _{x \rightarrow 0} \frac{\sqrt{\frac{1}{2}(1-\cos 2 x)}}{x}=\lim _{x \rightarrow 0} \frac{|\sin x|}{x}$
So, $\lim _{x \rightarrow 0^{+}} \frac{|\sin x|}{x}=1$ and $\lim _{x \rightarrow 0^{-}} \frac{|\sin x|}{x}=-1$
Hence, limit does not exist.
View full question & answer→MCQ 1222 Marks
If $f (x)=\left\{\begin{aligned} x \sin \frac{1}{x} ; & x \neq 0 \\ 0 ; & x=0\end{aligned}\right.$, then $\lim _{x \rightarrow 0} f (x)=$
Answer(B)
Here, $f(0)=0$
Since $-1 \leq \sin \frac{1}{x} \leq 1$
$\Rightarrow-x \leq x \sin \frac{1}{x} \leq x$
We know that, $\lim _{x \rightarrow 0}|x|=0$
$\therefore \quad \lim _{x \rightarrow 0} f(x)=0$
View full question & answer→MCQ 1232 Marks
$\lim _{x \rightarrow 0} \cos \frac{1}{x}$
Answer(C)
$\lim _{x \rightarrow 0} \cos \frac{1}{x}$ oscillates between -1 and 1 .
∴ Limit doesn't exist.
View full question & answer→MCQ 1242 Marks
$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals
- A
- B
$2 \sqrt{2}$
- ✓
$4 \sqrt{2}$
- D
$\sqrt{2}$
AnswerCorrect option: C. $4 \sqrt{2}$
(C)
$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$
$=\lim _{x \rightarrow 0}\left(\frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}\right)\left(\frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}-\sqrt{1+\cos x}}\right)$
$=\lim _{x \rightarrow 0}\left(\frac{\sin ^2 x}{1-\cos x}\right)(\sqrt{2}+\sqrt{1+\cos x})$
$=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{2 \sin ^2\left(\frac{x}{2}\right)}(\sqrt{2}+\sqrt{1+\cos x})$
$=\lim _{x \rightarrow 0} \frac{1}{2}\left(\frac{\sin x}{x}\right)^2 \times x^2 \times\left(\frac{\frac{x}{2}}{\sin \frac{x}{2}}\right)^2$ $\times \frac{1}{\left(\frac{x^2}{4}\right)} \times(\sqrt{2}+\sqrt{1+\cos x})$
$=\frac{1}{2} \times 4 \times 2 \sqrt{2}=4 \sqrt{2}$
View full question & answer→MCQ 1252 Marks
$\lim _{x \rightarrow 0} \frac{\sqrt{3+\cos x}-2}{\sin ^2 x}=$
- A
$\frac{1}{8}$
- B
$\frac{1}{4}$
- C
$-\frac{1}{4}$
- ✓
$-\frac{1}{8}$
AnswerCorrect option: D. $-\frac{1}{8}$
(D)
$\lim _{x \rightarrow 0} \frac{\sqrt{3+\cos x}-2}{\sin ^2 x}$
$=\lim _{x \rightarrow 0} \frac{(3+\cos x)-4}{\sin ^2 x(\sqrt{3+\cos x}+2)}$
$=\lim _{x \rightarrow 0} \frac{-(1-\cos x)}{(1-\cos x)(1+\cos x)(\sqrt{3+\cos x}+2)}=-\frac{1}{8}$
View full question & answer→MCQ 1262 Marks
$\lim _{h \rightarrow 0} \frac{2\left[\sqrt{3} \sin \left(\frac{\pi}{6}+h\right)-\cos \left(\frac{\pi}{6}+h\right)\right]}{\sqrt{3} h(\sqrt{3} \cos h-\sin h)}=$
- A
$-\frac{2}{3}$
- B
$-\frac{3}{4}$
- C
$-2 \sqrt{3}$
- ✓
$\frac{4}{3}$
AnswerCorrect option: D. $\frac{4}{3}$
(D)
$\lim _{ h \rightarrow 0} \frac{2\left[\sqrt{3} \sin \left(\frac{\pi}{6}+ h \right)-\cos \left(\frac{\pi}{6}+ h \right)\right]}{\sqrt{3} h(\sqrt{3} \cos h-\sin h )}$
$=\lim _{h \rightarrow 0} \frac{\frac{4}{\sqrt{3}}\left[\frac{\sqrt{3}}{2} \sin \left(\frac{\pi}{6}+h\right)-\frac{1}{2} \cos \left(\frac{\pi}{6}+h\right)\right]}{h(\sqrt{3} \cos h-\sin h)}$
$=\lim _{h \rightarrow 0} \frac{\frac{4}{\sqrt{3}} \sin \left[\left(\frac{\pi}{6}+h\right)-\frac{\pi}{6}\right]}{h(\sqrt{3} \cos h-\sin h)}$
$=\frac{4}{\sqrt{3}} \lim _{ h \rightarrow 0} \frac{\sin h}{ h } \cdot \frac{1}{(\sqrt{3} \cos h-\sin h )}$
$=\frac{4}{\sqrt{3}} \cdot 1 \cdot \frac{1}{(\sqrt{3}-0)}$
$=\frac{4}{3}$
View full question & answer→MCQ 1272 Marks
$\lim _{y \rightarrow 0} \frac{(x+y) \sec (x+y)-x \sec x}{y}=$
- ✓
$\sec x(x \tan x+1)$
- B
$x \tan x+\sec x$
- C
$x \sec x+\tan x$
- D
AnswerCorrect option: A. $\sec x(x \tan x+1)$
(A)
$\lim _{y \rightarrow 0} \frac{(x+y) \sec (x+y)-x \sec x}{y}$
$=\lim _{y \rightarrow 0}\left\{\frac{x[\sec (x+y)-\sec x]}{y}+\sec (x+y)\right\}$
$=\lim _{y \rightarrow 0}\left[\frac{x}{y} \cdot \frac{\cos x-\cos (x+y)}{\cos (x+y) \cos x}\right]+\lim _{y \rightarrow 0} \sec (x+y)$
$=\lim _{y \rightarrow 0}\left\lfloor\frac{x}{y} \cdot \frac{2 \sin \left(x+\frac{y}{2}\right) \sin \left(\frac{y}{2}\right)}{\cos (x+y) \cos x}\right\rfloor+\sec x$
$=\lim _{y \rightarrow 0}\left[\frac{x \sin \left(x+\frac{y}{2}\right)}{\cos (x+y) \cdot \cos x} \cdot \frac{\sin \left(\frac{y}{2}\right)}{\frac{y}{2}}\right]+\sec x$
$=\left(\frac{x \sin x}{\cos x \cdot \cos x} \cdot 1\right)+\sec x$
$=x \tan x \sec x+\sec x=\sec x(x \tan x+1)$
View full question & answer→MCQ 1282 Marks
$\lim _{x \rightarrow 0} \frac{x^2(\tan 2 x-2 \tan x)^2}{(1-\cos 2 x)^4}=$
- A
- B
- C
$\frac{1}{2}$
- ✓
$\frac{1}{4}$
AnswerCorrect option: D. $\frac{1}{4}$
(D)
$\lim _{x \rightarrow 0} \frac{x^2(\tan 2 x-2 \tan x)^2}{(1-\cos 2 x)^4}$
$=\lim _{x \rightarrow 0} \frac{x^2\left(\frac{2 \tan x}{1-\tan ^2 x}-2 \tan x\right)^2}{\left(2 \sin ^2 x\right)^4}$
$=\lim _{x \rightarrow 0} \frac{x^2\left(2 \tan ^3 x\right)^2}{\left(1-\tan ^2 x\right)^2\left(2 \sin ^2 x\right)^4}$
$=\lim _{x \rightarrow 0} \frac{4 x^2 \tan ^6 x}{16\left(1-\tan ^2 x\right)^2 \sin ^8 x}$
$=\lim _{x \rightarrow 0} \frac{1}{4\left(1-\tan ^2 x\right)^2 \cos ^6 x\left(\frac{\sin x}{x}\right)^2}$
$=\frac{1}{4(1-0)^2(1)(1)^2}$
$=\frac{1}{4}$
View full question & answer→MCQ 1292 Marks
$\lim _{\theta \rightarrow 0} \frac{4 \theta(\tan \theta-\sin \theta)}{(1-\cos 2 \theta)^2}$ is
- A
$\frac{1}{\sqrt{2}}$
- ✓
$\frac{1}{2}$
- C
- D
AnswerCorrect option: B. $\frac{1}{2}$
(B)
$\lim _{\theta \rightarrow 0} \frac{4 \theta(\tan \theta-\sin \theta)}{(1-\cos 2 \theta)^2}=4 \lim _{\theta \rightarrow 0} \frac{\theta(\tan \theta-\sin \theta)}{\left(2 \sin ^2 \theta\right)^2}$
$=4 \lim _{\theta \rightarrow 0} \frac{\theta \sin \theta(1-\cos \theta)}{4 \sin ^4 \theta \cos \theta}$
$=\lim _{\theta \rightarrow 0}\left(\frac{\theta}{\sin \theta}\right) \frac{2 \sin ^2 \frac{\theta}{2}}{\sin ^2 \theta \cos \theta}$
$=\lim _{\theta \rightarrow 0}\left(\frac{\theta}{\sin \theta}\right) \frac{2 \sin ^2 \frac{\theta}{2}}{\left[2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)\right]^2} \cdot \frac{1}{\cos \theta}$
$=\frac{1}{2} \lim _{\theta \rightarrow 0}\left(\frac{\theta}{\sin \theta}\right) \frac{1}{\cos ^2\left(\frac{\theta}{2}\right) \cdot \cos \theta}=\frac{1}{2}$
View full question & answer→MCQ 1302 Marks
$\lim _{x \rightarrow 0} \frac{\cos (\sin x)-1}{x^2}=$
- A
- B
- C
$\frac{1}{2}$
- ✓
$-\frac{1}{2}$
AnswerCorrect option: D. $-\frac{1}{2}$
(D)
$\lim _{x \rightarrow 0} \frac{\cos (\sin x)-1}{x^2}=\lim _{x \rightarrow 0} \frac{-2 \sin ^2\left(\frac{\sin x}{2}\right)}{x^2}$
$=-2 \lim _{x \rightarrow 0} \frac{\sin ^2\left(\frac{\sin x}{2}\right)}{\frac{\sin ^2 x}{4}} \times \frac{\sin ^2 x}{4 x^2}$
$=-\frac{1}{2} \lim _{x \rightarrow 0}\left[\frac{\sin \left(\frac{\sin x}{2}\right)}{\frac{\sin x}{2}}\right]^2 \times\left(\frac{\sin x}{x}\right)^2$
$=-\frac{1}{2}(1)^2(1)^2=\frac{-1}{2}$
View full question & answer→MCQ 1312 Marks
$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}$ is equal to
- A
- ✓
- C
$\frac{-1}{4}$
- D
$\frac{1}{2}$
Answer(B)
$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}$
$=\lim _{x \rightarrow 0} \frac{2 \sin ^2 x(3+\cos x)}{x \tan 4 x}$
$=2 \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^2 \times\left(\frac{1}{\frac{\tan 4 x}{4 x}}\right) \times \frac{1}{4} \times(3+\cos x)=2$
View full question & answer→MCQ 1322 Marks
$\lim _{x \rightarrow 0} \frac{x^3 \cot x}{1-\cos x}=$
Answer(C)
$\lim _{x \rightarrow 0} \frac{x^3 \cot x}{1-\cos x}=\lim _{x \rightarrow 0}\left(\frac{x^3 \cot x}{1-\cos x} \times \frac{1+\cos x}{1+\cos x}\right)$
$=\lim _{x \rightarrow 0} \frac{\left(x^3 \cot x\right)(1+\cos x)}{1-\cos ^2 x}$
$=\lim _{x \rightarrow 0} \frac{\left(x^3 \cot x\right)(1+\cos x)}{\sin ^2 x}$
$=\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^3 \times \lim _{x \rightarrow 0} \cos x \times \lim _{x \rightarrow 0}(1+\cos x)$
$=(1)^3(1)(1+1)=2$
View full question & answer→MCQ 1332 Marks
The value of $\lim _{x \rightarrow 0} \frac{\sin a-\tan a}{\sin ^3 a}$ will be
- ✓
$-\frac{1}{2}$
- B
$\frac{1}{2}$
- C
- D
AnswerCorrect option: A. $-\frac{1}{2}$
(A)
$\lim _{a \rightarrow 0} \frac{\sin a-\tan a}{\sin ^3 a}=\lim _{a \rightarrow 0} \frac{\sin a-\frac{\sin a}{\cos a}}{\sin ^3 a}$
$=\lim _{a \rightarrow 0} \frac{\cos a-1}{\sin ^2 a \cos a}$
$=\lim _{a \rightarrow 0} \frac{-(1-\cos a)}{\left(1-\cos ^2 a\right)(\cos a)}$
$=\lim _{a \rightarrow 0}\left[-\frac{1}{(1+\cos a) \cos a}\right]=\frac{-1}{2}$
View full question & answer→MCQ 1342 Marks
$\lim _{x \rightarrow 0} \frac{x \cos x-\sin x}{x^2 \sin x}=$
AnswerCorrect option: B. $-\frac{1}{3}$
(B)
$\lim _{x \rightarrow 0} \frac{x \cos x-\sin x}{x^2 \sin x}$
$=\lim _{x \rightarrow 0} \frac{-\sin x}{2 \sin x+x \cos x} \ldots[$ [By L-Hospital's rule] $]$
$=\lim _{x \rightarrow 0} \frac{-\cos x}{3 \cos x-x \sin x}$
...[Again by L-Hospital's rule]
$=-\frac{1}{3}$
View full question & answer→MCQ 1352 Marks
$\lim _{x \rightarrow 0} \frac{1-\cos x-\cos 2 x+\cos x \cos 2 x}{x^4}$ is equal to
- ✓
- B
- C
$\frac{1}{3}$
- D
$\frac{1}{2}$
Answer(A)
$\lim _{x \rightarrow 0} \frac{1-\cos x-\cos 2 x+\cos x \cdot \cos 2 x}{x^4}$
$=\lim _{x \rightarrow 0} \frac{(1-\cos x)-\cos 2 x(1-\cos x)}{x^4}$
$=\lim _{x \rightarrow 0} \frac{(1-\cos x)(1-\cos 2 x)}{x^4}$
$=\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^2}\right)\left(\frac{1-\cos 2 x}{x^2}\right)$
$\lim _{x \rightarrow 0} \frac{1-\cos k x}{x^2}=\frac{ k ^2}{2}$
$=\left(\frac{1^2}{2}\right)\left(\frac{2^2}{2}\right)=1 $
View full question & answer→MCQ 1362 Marks
$\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$ equals
- A
$-\pi$
- ✓
$\pi$
- C
$\frac{\pi}{2}$
- D
Answer(B)
$\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}=\lim _{x \rightarrow 0} \frac{\sin \left(\pi-\pi \cos ^2 x\right)}{x^2}$
$=\lim _{x \rightarrow 0} \frac{\sin \left(\pi \sin ^2 x\right)}{\pi \sin ^2 x} \times \frac{\pi \sin ^2 x}{x^2}=\pi$
View full question & answer→MCQ 1372 Marks
$\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{\cos 2 x-\cos 8 x}=$
- A
$\frac{1}{17}$
- ✓
$\frac{1}{5}$
- C
$\frac{1}{15}$
- D
$\frac{1}{7}$
AnswerCorrect option: B. $\frac{1}{5}$
(B)
$\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{\cos 2 x-\cos 8 x}$
$=\lim _{x \rightarrow 0} \frac{2 \sin ^2 x}{2 \sin \left(\frac{2 x+8 x}{2}\right) \sin \left(\frac{8 x-2 x}{2}\right)}$
$=\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sin 5 x \cdot \sin 3 x}=\frac{1}{15}$
View full question & answer→MCQ 1382 Marks
$\lim _{x \rightarrow 0} \frac{\sin (2+x)-\sin (2-x)}{x}=$
- A
$\sin 2$
- B
$2 \sin 2$
- ✓
$2 \cos 2$
- D
AnswerCorrect option: C. $2 \cos 2$
(C)
$\lim _{x \rightarrow 0} \frac{\sin (2+x)-\sin (2-x)}{x}=\lim _{x \rightarrow 0} \frac{2 \cos 2 \cdot \sin x}{x}$
$=2 \cos 2 \lim _{x \rightarrow 0} \frac{\sin x}{x}$
$=2 \cos 2$
View full question & answer→MCQ 1392 Marks
The value of $\lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta}$ is
- ✓
$\frac{4}{9}$
- B
$\frac{9}{4}$
- C
$\frac{9}{3}$
- D
$\frac{3}{4}$
AnswerCorrect option: A. $\frac{4}{9}$
(A)
Applying L-Hospital's rule 2 times, we get
$\lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta}=\lim _{\theta \rightarrow 0} \frac{4 \sin 4 \theta}{6 \sin 6 \theta}$
$=\lim _{\theta \rightarrow 0} \frac{16 \cos 4 \theta}{36 \cos 6 \theta}=\frac{4}{9}$
View full question & answer→MCQ 1402 Marks
$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=$
- A
$\frac{m}{n}$
- B
$\frac{ n }{ m }$
- ✓
$\frac{ m ^2}{ n ^2}$
- D
$\frac{ n ^2}{m^2}$
AnswerCorrect option: C. $\frac{ m ^2}{ n ^2}$
(C)
$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0}\left\{\frac{2 \sin ^2 \frac{m x}{2}}{2 \sin ^2 \frac{n x}{2}}\right\}$
$=\lim _{x \rightarrow 0}\left[\left\{\frac{\sin \frac{m x}{2}}{\frac{m x}{2}}\right\}^2 \frac{m^2 x^2}{4} \cdot \frac{1}{\left\{\frac{\sin \frac{n x}{2}}{\frac{n x}{2}}\right\}^2} \cdot \frac{4}{n^2 x^2}\right]$
$=\frac{ m ^2}{ n ^2} \times l =\frac{ m ^2}{ n ^2}$
Alternate method:
Applying L-Hospital's rule 2 times, we get
$\lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}=\lim _{x \rightarrow 0} \frac{m \sin m x}{n \sin n x}$
$=\lim _{x \rightarrow 0} \frac{m^2 \cos m x}{n^2 \cos n x}=\frac{m^2}{n^2}$
View full question & answer→MCQ 1412 Marks
The value of $\lim _{x \rightarrow 0} \frac{\sin ^2 x+\cos x-1}{x^2}$ is
- A
- ✓
$\frac{1}{2}$
- C
$-\frac{1}{2}$
- D
$0$
AnswerCorrect option: B. $\frac{1}{2}$
(B)
$\lim _{x \rightarrow 0} \frac{\sin ^2 x+\cos x-1}{x^2}=\lim _{x \rightarrow 0} \frac{\cos x-\cos ^2 x}{x^2}$
$=\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^2}\right) \cos x$
$\lim _{x \rightarrow 0} \frac{1-\cos k x}{x^2}=\frac{ k ^2}{2}$
$=\frac{1^2}{2} \times 1=\frac{1}{2}$
View full question & answer→MCQ 1422 Marks
$\lim _{x \rightarrow 0} \frac{\sin x^2\left(1-\cos x^2\right)}{x^6}$ is equal to
- A
- B
$\sqrt{2}$
- ✓
$\frac{1}{2}$
- D
AnswerCorrect option: C. $\frac{1}{2}$
(C)
$\lim _{x \rightarrow 0} \frac{\sin x^2\left(1-\cos x^2\right)}{x^6}=\lim _{x \rightarrow 0} \frac{\sin x^2}{x^2} \cdot \frac{1-\cos x^2}{x^4}$
$=\lim _{x \rightarrow 0} \frac{1-\cos x^2}{x^4}$
$=\lim _{x \rightarrow 0} \frac{\sin x^2 \cdot 2 x}{4 x^3}$
$=\lim _{x \rightarrow 0} \frac{\sin x^2}{2 x^2}=\frac{1}{2}$
View full question & answer→MCQ 1432 Marks
$\lim _{x \rightarrow 0} \frac{1-\cos x}{x \sin x}$ is equal to
AnswerCorrect option: D. $\frac{1}{2}$
(D)
$\lim _{x \rightarrow 0} \frac{1-\cos x}{x \sin x}=\lim _{x \rightarrow 0} \frac{1-\cos ^2 x}{x \sin x(1+\cos x)}$
$=\lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \frac{1}{1+\cos x}$
$=1 \cdot \frac{1}{1+1}=\frac{1}{2}$
View full question & answer→MCQ 1442 Marks
If $\lim _{x \rightarrow 0} \frac{\sin p x}{\tan 3 x}=4$, then the value of $p$ is
Answer(C)
$\lim _{x \rightarrow 0} \frac{\sin p x}{\tan 3 x}=\lim _{x \rightarrow 0} \frac{\frac{\sin p x}{ p x} \times p }{\frac{\tan 3 x}{3 x} \times 3}=\frac{1( p )}{1(3)}=\frac{ p }{3}$
Since $\lim _{x \rightarrow 0} \frac{\sin p x}{\tan 3 x}=4$
$\therefore \quad \frac{p}{3}=4 \Rightarrow p=12$
View full question & answer→MCQ 1452 Marks
$\lim _{x \rightarrow 0} \frac{\sin 2 x+\sin 6 x}{\sin 5 x-\sin 3 x}=$
- A
$\frac{1}{2}$
- B
$\frac{1}{4}$
- C
- ✓
Answer(D)
$\lim _{x \rightarrow 0} \frac{2 \sin 4 x \cos 2 x}{2 \sin x \cos 4 x}$
$=\lim _{x \rightarrow 0} 4\left(\frac{\sin 4 x}{4 x}\right)\left(\frac{x}{\sin x}\right) \frac{\cos 2 x}{\cos 4 x}=4$
Alternate method:
$\lim _{x \rightarrow 0} \frac{\frac{2 \sin 2 x}{2 x}+\frac{6 \sin 6 x}{6 x}}{\frac{5 \sin 5 x}{5 x}-\frac{3 \sin 3 x}{3 x}}=\frac{2+6}{5-3}=4$
View full question & answer→MCQ 1462 Marks
$\lim _{x \rightarrow 0} \frac{4 \sin ^2 x+2 x \cos x}{3 x+\tan 3 x}=$
- A
$\frac{2}{7}$
- B
$\frac{11}{3}$
- C
$\frac{13}{7}$
- ✓
$\frac{1}{3}$
AnswerCorrect option: D. $\frac{1}{3}$
(D)
$\lim _{x \rightarrow 0} \frac{4 \sin ^2 x+2 x \cos x}{3 x+\tan 3 x}$
$=\frac{4 \lim _{x \rightarrow 0} \frac{\sin x}{x} \times \sin x+2 \lim _{x \rightarrow 0} \frac{x \cos x}{x}}{\lim _{x \rightarrow 0} \frac{3 x}{x}+\lim _{x \rightarrow 0} \frac{\tan 3 x}{x}}=\frac{1}{3}$
View full question & answer→MCQ 1472 Marks
The value of $\lim _{x \rightarrow 2 a} \frac{\sqrt{x-2 a}+\sqrt{x}-\sqrt{2 a}}{\sqrt{x^2-4 a^2}}$ is
- A
$\frac{1}{\sqrt{a}}$
- ✓
$\frac{1}{2 \sqrt{a}}$
- C
$\frac{\sqrt{a}}{2}$
- D
$2 \sqrt{a}$
AnswerCorrect option: B. $\frac{1}{2 \sqrt{a}}$
(B)
$\lim _{x \rightarrow 2 a } \frac{\sqrt{x-2 a }+\sqrt{x}-\sqrt{2 a }}{\sqrt{x^2-4 a ^2}}$
$=\lim _{x \rightarrow 2 a } \frac{\sqrt{x-2 a }+\frac{x-2 a }{\sqrt{x}+\sqrt{2 a }}}{\sqrt{(x-2 a )} \sqrt{(x+2 a )}}$
$=\lim _{x \rightarrow 2 a }\left(\frac{1}{\sqrt{x+2 a }}+\frac{\sqrt{x-2 a }}{\sqrt{(x+2 a )}(\sqrt{x}+\sqrt{2 a })}\right)$
$=\frac{1}{\sqrt{4 a}}=\frac{1}{2 \sqrt{a}}$
View full question & answer→MCQ 1482 Marks
The value of $\lim _{x \rightarrow 2} \frac{\sqrt{1+\sqrt{2}+x-\sqrt{3}}}{x-2}$ is
- ✓
$\frac{1}{8 \sqrt{3}}$
- B
$\frac{1}{\sqrt{3}}$
- C
$8 \sqrt{3}$
- D
$\sqrt{3}$
AnswerCorrect option: A. $\frac{1}{8 \sqrt{3}}$
(A)
$\lim _{x \rightarrow 2} \frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}$
$=\lim _{x \rightarrow 2} \frac{1+\sqrt{2+x}-3}{(x-2)(\sqrt{1+\sqrt{2+x}}+\sqrt{3})}$
$=\lim _{x \rightarrow 2} \frac{\sqrt{2+x}-2}{(x-2)(\sqrt{1+\sqrt{2+x}}+\sqrt{3})}$
$=\lim _{x \rightarrow 2} \frac{1}{(\sqrt{1+\sqrt{2+x}}+\sqrt{3})(\sqrt{2+x}+2)}$
$=\frac{1}{2 \sqrt{3} \times 4}=\frac{1}{8 \sqrt{3}}$
View full question & answer→MCQ 1492 Marks
The value of $\lim _{x \rightarrow 0} \frac{\sqrt{x^2+1}-1}{\sqrt{x^2+9}-3}$ is
Answer(A)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{\sqrt{x^2+1}-1}{\sqrt{x^2+9}-3}=\lim _{x \rightarrow 0} \frac{\frac{2 x}{2 \sqrt{x^2+1}}}{\frac{2 x}{2 \sqrt{x^2+9}}}$
$=\lim _{x \rightarrow 0} \frac{\sqrt{x^2+9}}{\sqrt{x^2+1}}=3$
View full question & answer→MCQ 1502 Marks
$\lim _{x \rightarrow 4} \frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=$
- A
$0$
- B
$\frac{1}{3}$
- ✓
$-\frac{1}{3}$
- D
AnswerCorrect option: C. $-\frac{1}{3}$
(C)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 4} \frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\lim _{x \rightarrow 4} \frac{\frac{1}{2 \sqrt{5+x}}}{-\frac{1}{2 \sqrt{5-x}}}$
$=-\lim _{x \rightarrow 4} \frac{\sqrt{5-x}}{\sqrt{5+x}}=\frac{-1}{\sqrt{9}}=-\frac{1}{3}$
View full question & answer→