Question

lim x tends to pie /2[cosec x-1/(pie /2-x)^2]​

Answer 1

Given to evaluate,

$\displaystyle\sf{\longrightarrow L=\lim_{x\to\frac {\pi}{2}}\dfrac {\csc x-1}{\left (\dfrac {\pi}{2}-x\right)^2}}$

$\displaystyle\sf{\longrightarrow L=\lim_{x\to\frac {\pi}{2}}\dfrac {\dfrac {1}{\sin x}-1}{\left (\dfrac {\pi}{2}-x\right)^2}}$

$\displaystyle\sf{\longrightarrow L=\lim_{x\to\frac {\pi}{2}}\dfrac {\dfrac {1}{\sin\left (\dfrac {\pi}{2}+x-\dfrac {\pi}{2}\right)}-1}{\left (x-\dfrac {\pi}{2}\right)^2}\quad\quad\dots(1)}$

Substitute,

$\displaystyle\sf{\longrightarrow u=x-\dfrac {\pi}{2}}$

Then we see,

$\displaystyle\sf{\longrightarrow x\to\dfrac {\pi}{2}\quad\implies\quad u\to0}$

Then (1) becomes,

$\displaystyle\sf{\longrightarrow L=\lim_{u\to0}\dfrac {\dfrac {1}{\sin\left (\dfrac {\pi}{2}+u\right)}-1}{u^2}}$

$\displaystyle\sf{\longrightarrow L=\lim_{u\to0}\dfrac {\dfrac {1}{\cos u}-1}{u^2}}$

$\displaystyle\sf{\longrightarrow L=\lim_{u\to0}\dfrac {1-\cos u}{u^2\cos u}}$

$\displaystyle\sf{\longrightarrow L=\lim_{u\to0}\dfrac {2\sin^2\left (\dfrac {u}{2}\right)}{4\cdot\dfrac {u^2}{4}\cos u}}$

$\displaystyle\sf{\longrightarrow L=\lim_{u\to0}\dfrac {2}{4\cos u}\cdot\dfrac {\sin^2\left (\dfrac {u}{2}\right)}{\dfrac {u^2}{4}}}$

$\displaystyle\sf{\longrightarrow L=\lim_{u\to0}\dfrac {1}{2\cos u}\cdot\lim_{u\to0}\dfrac {\sin^2\left (\dfrac {u}{2}\right)}{\left (\dfrac {u}{2}\right)^2}}$

Since $\displaystyle\sf {\lim_{t\to0}\dfrac {\sin t}{t}=1,}$

$\displaystyle\sf{\longrightarrow L=\dfrac {1}{2\cos 0}\cdot1^2}$

$\displaystyle\sf{\longrightarrow \underline {\underline {L=\dfrac {1}{2}}}}$