Question

Evaluate the limit

$lim \: x \: \to \: y \: \frac{sinx - siny}{x - y}$

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Answer 1

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

Given expression is

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

If we substitute directly x = y, we get

$\rm \:  =  \: \dfrac{siny - siny}{y - y}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form.

So, Consider again

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

$\rm \:  =  \: \displaystyle\lim_{x \to y} \frac{2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}{x - y}$

$\rm \:  =  \: \displaystyle\lim_{x \to y} 2cos\bigg[\dfrac{x + y}{2} \bigg] \times \displaystyle\lim_{x \to y} \frac{sin\bigg[\dfrac{x - y}{2} \bigg]}{x - y}$

$\rm \:  =  \: 2cos\bigg[\dfrac{y + y}{2} \bigg] \times \displaystyle\lim_{x - y\to 0} \frac{sin\bigg[\dfrac{x - y}{2} \bigg]}{\dfrac{x - y}{2} \times 2}$

$\rm \:  =  \: cos\bigg[\dfrac{2y}{2} \bigg] \times \displaystyle\lim_{x - y\to 0} \frac{sin\bigg[\dfrac{x - y}{2} \bigg]}{\dfrac{x - y}{2}}$

We know

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

$\rm \:  =  \: cosy \times 1$

$\rm \:  =  \: cosy$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to y} \frac{sinx - siny}{x - y} \: = \: cosy \: }}}$

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More to Know

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

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