Question

with explanation plz

Answer 1

$\large\underline{\sf{Given \:Question - }}$

$\rm :\longmapsto\:\displaystyle\lim_{x \to \pi} \frac{sinx}{x - \pi}$

(A) 1

(B) 2

(C) - 1

(D) - 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to \pi} \frac{sinx}{x - \pi}$

If we put directly the value of x, we get

$\rm \:  =  \: \dfrac{sin\pi}{\pi - \pi}$

$\rm \:  =  \: \dfrac{0}{0} \: \: which \: is \: meaningless$

So, to evaluate this limit,

$\rm :\longmapsto\:\displaystyle\lim_{x \to \pi} \frac{sinx}{x - \pi}$

$\red{ \boxed{ \sf{ \:Put \: x = \pi - h, \: as \: x \: \to \: \pi, \: so \: h \: \to \: 0}}}$

So, we get

$\rm \:  =  \: \displaystyle\lim_{h \to 0} \frac{sin(\pi - h)}{\pi - h - \pi}$

$\rm \:  =  \: \displaystyle\lim_{h \to 0} \frac{sinh}{ - h}$

$\red{\bigg \{ \because \:sin(\pi - h) = - \: sinh \bigg \}}$

$\rm \:  =  - \: \displaystyle\lim_{h \to 0} \frac{sinh}{ h}$

$\rm \:  =  \: - \: 1$

Hence,

$\rm :\longmapsto\:\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to \pi} \frac{sinx}{x - \pi} = - \: 1}}}$

• Option (C) is correct.

Additional Information :-

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \frac{sinx}{x} = 1 \: \: }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \frac{tanx}{x} = 1 \: \: }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \frac{log(1 + x)}{x} = 1 \: \: }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \frac{ {e}^{x} - 1 }{x} = 1 \: \: }}}$

$\red{ \boxed{ \sf{ \:\displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1 }{x} = loga \: \: }}}$