Displaystyle lim-xrightarrow 1-1-x-tan-displaystyle frac-pi x-2-pi 2pidisplaystyle frac-pi-2-displaystyle frac-2-pi-

Question

Toppr Admin · Accepted Answer

Limx-x2192-1-1-x2212-x-tan-x3C0-x2-limx-x2192-11-x2212-xcot-x3C0-x2-It is of the form 00- so applying L-Hospital-apos-s rule-xA0-limx-x2192-1-x2212-1-x2212-x3C0-2csc2-x3C0-x2-2-x3C0-

$lim x \to 1 (1 - x) tan (\frac{π x}{2}) =$
$π$
$2 π$
$\frac{π}{2}$
$\frac{2}{π}$

$lim x \to 1 (1 - x) tan (\frac{π x}{2})$
$= {lim}_{x \to 1} \frac{1 - x}{cot (\frac{π x}{2})}$
It is of the form $\frac{0}{0}$ , so applying L-Hospital's rule
$= {lim}_{x \to 1} \frac{- 1}{- \frac{π}{2} {csc}^{2} (\frac{π x}{2})}$
$= \frac{2}{π}$

limx→1(1−x)tan(πx2)=π2ππ22π

limx→1(1−x)tan(πx2)=limx→11−xcot(πx2)It is of the form 00, so applying L-Hospital's rule =limx→1−1−π2csc2(πx2)=2π

$lim x \to 1 (1 - x) tan (\frac{π x}{2}) =$
$π$
$2 π$
$\frac{π}{2}$
$\frac{2}{π}$

$lim x \to 1 (1 - x) tan (\frac{π x}{2})$
$= {lim}_{x \to 1} \frac{1 - x}{cot (\frac{π x}{2})}$
It is of the form $\frac{0}{0}$ , so applying L-Hospital's rule
$= {lim}_{x \to 1} \frac{- 1}{- \frac{π}{2} {csc}^{2} (\frac{π x}{2})}$
$= \frac{2}{π}$