Question

lim (Sin(pi -x)) / (pi-x) x-> pi

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to \pi}\sf \frac{sin(\pi - x)}{\pi - x}$

If we substitute directly the value of x, we get

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

$\rm \: = \: \dfrac{sin0}{0}$

$\rm \: = \: \dfrac{0}{0}$

which is indeterminant form.

So, to solve this question, we use Method of Substitution

$\rm :\longmapsto\:\displaystyle\lim_{x \to \pi}\sf \frac{sin(\pi - x)}{\pi - x}$

Now, Substitute

$\red{\rm :\longmapsto\:x = \pi - y, \: \: as\: x \to \: \pi, \: so \: y \: \to \: 0 \: }$

So, above expression can be rewritten as

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

$\rm \: = \: \displaystyle\lim_{y \to 0}\sf \frac{siny}{y}$

$\rm \: = \: 1$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to \pi}\sf \frac{sin(\pi - x)}{\pi - x} = 1 \: }}$

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Alternative Method

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to \pi}\sf \frac{sin(\pi - x)}{\pi - x}$

can be rewritten as

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

We know,

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

So, using this

$\rm \: = \: 1$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to \pi}\sf \frac{sin(\pi - x)}{\pi - x} = 1 \: }}$

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Additional Information :-

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

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

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

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

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