Question

Evaluate the following

$lim_{x \: \to \: 0}( \frac{(x + y)sin(x + y) - ysiny}{x} )$
​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{(x + y)sin(x + y) - ysiny}{x}$

If we substitute directly x = 0, we get

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

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form.

Consider, again

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{(x + y)sin(x + y) - ysiny}{x}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{xsin(x + y) + ysin(x + y) - ysiny}{x}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{xsin(x + y) }{x} + \displaystyle\lim_{x \to 0}\rm \dfrac{ysin(x + y) - ysiny}{x}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm sin(x + y) + y\displaystyle\lim_{x \to 0}\rm \dfrac{sin(x + y) - siny}{x}$

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

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

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

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

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

We know,

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

So, using this, we get

$\rm \:  =  \: siny + y \times 1 \times cosy$

$\rm \:  =  \: siny + y cosy$

Hence,

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{(x + y)sin(x + y) - ysiny}{x} = siny + y \: cosy}}$

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LEARN MORE

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