The value of mathop {lim }limits_{x to frac{p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We know that $\frac{1 - 	an(x/2)}{1 + 	an(x/2)} = 	an(\frac{\pi}{4} - \frac{x}{2})$.So,the limit becomes $\mathop {\lim }\limits_{x 	o \frac{\

Vedclass · Accepted Answer

$\frac{1}{32}$

Vedclass · Answer

(C) We know that $\frac{1 - 	an(x/2)}{1 + 	an(x/2)} = 	an(\frac{\pi}{4} - \frac{x}{2})$.So,the limit becomes $\mathop {\lim }\limits_{x 	o \frac{\pi }{2}} \frac{{	an \left( {\frac{\pi }{4} - \frac{x}{2}} ight)\,(1 - \sin x)}}{{{{(\pi - 2x)}^3}}}$.Let $x = \frac{\pi }{2} + y$,where $y 	o 0$. Then $\pi - 2x = \pi - 2(\frac{\pi}{2} + y) = -2y$.Also,$1 - \sin x = 1 - \sin(\frac{\pi}{2} + y) = 1 - \cos y = 2\sin^2(\frac{y}{2})$.Substituting these,we get $\mathop {\lim }\limits_{y 	o 0} \frac{{	an \left( {\frac{\pi}{4} - (\frac{\pi}{4} + \frac{y}{2})} ight)\,(2\sin^2 \frac{y}{2})}}{{{{( - 2y)}^3}}}$.$= \mathop {\lim }\limits_{y 	o 0} \frac{{-	an(\frac{y}{2}) \cdot 2\sin^2(\frac{y}{2})}}{-8y^3} = \mathop {\lim }\limits_{y 	o 0} \frac{1}{4} \cdot \frac{	an(y/2)}{y} \cdot \left(\frac{\sin(y/2)}{y}ight)^2$.$= \frac{1}{4} \cdot \frac{1}{2} \cdot (\frac{1}{2})^2 = \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{1}{32}$.

The value of $\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\left[ {1 - \tan \left( {\frac{x}{2}} \right)} \right]\,[1 - \sin x]}}{{\left[ {1 + \tan \left( {\frac{x}{2}} \right)} \right]\,{{[\pi - 2x]}^3}}}$ is

Similar Questions

$\lim _{x \rightarrow 1} (1 + \log _{e} x)^{1 / \log _{e} x}$ is equal to

If $l, m$ $(l < m)$ are roots of $ax^2 + bx + c = 0$,then $\lim_{x \rightarrow \alpha} \frac{|ax^2 + bx + c|}{ax^2 + bx + c} = $

If $\mathop {\lim }\limits_{x \to 5} \frac{{{x^k} - {5^k}}}{{x - 5}} = 500$,then the positive integral value of $k$ is

If $f(x) = \frac{5x \operatorname{cosec}(\sqrt{x}) - 1}{(x - 2) \operatorname{cosec}(\sqrt{x})}$,then $\lim_{x \rightarrow \infty} f(x^2) = $

Find $\mathop {\lim }\limits_{x \to 0} f(x)$ and $\mathop {\lim }\limits_{x \to 1} f(x),$ where $f(x) = \begin{cases} 2x+3, & x \leq 0 \\ 3(x+1), & x > 0 \end{cases}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module