Question

Evaluate the limit. ​

Answer 1

Given Question :-

Evaluate the following limit :

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{log(1 + sin4x)}{ {e}^{sin5x} - 1}$

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

Consider,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{log(1 + sin4x)}{ {e}^{sin5x} - 1}$

If we put directly x = 0, we get

$\rm \:  =  \:\dfrac{log(1 + sin0)}{ {e}^{sin0} - 1}$

$\rm \:  =  \:\dfrac{log(1 +0)}{ {e}^{0} - 1}$

$\rm \:  =  \:\dfrac{log(1)}{1 - 1}$

$\rm \:  =  \:\dfrac{0}{0}$

which is indeterminant form

So,

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{log(1 + sin4x)}{ {e}^{sin5x} - 1}$

can be rewritten as

$\rm \:  =  \:\displaystyle\lim_{x \to 0} \frac{log(1 + sin4x)}{sin4x} \times \frac{sin5x}{ {e}^{sin5x} - 1} \times \frac{sin4x}{sin5x}$

can be further rewritten as

$\rm=\displaystyle\lim_{x \to 0} \frac{log(1 + sin4x)}{sin4x} \times \displaystyle\lim_{x \to 0} \frac{sin5x}{ {e}^{sin5x} - 1} \times \displaystyle\lim_{x \to 0} \frac{sin4x}{sin5x}$

We know,

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

and

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

So, using these Identities, we get

$\rm \:  =  \:1 \times 1 \times \displaystyle\lim_{x \to 0} \frac{sin4x}{sin5x}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0} \frac{sin4x}{sin5x}$

can be rewritten as

$\rm \:  =  \:\displaystyle\lim_{x \to 0} \frac{sin4x}{4x} \times \frac{5x}{sin5x} \times \frac{4x}{5x}$

$\rm \:  =  \:\displaystyle\lim_{x \to 0} \frac{sin4x}{4x} \times \displaystyle\lim_{x \to 0} \frac{5x}{sin5x} \times \frac{4}{5}$

We know,

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

So, using this identity, we get

$\rm \:  =  \:1 \times 1\times \dfrac{4}{5}$

$\rm \:  =  \: \dfrac{4}{5}$

Hence,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{log(1 + sin4x)}{ {e}^{sin5x} - 1} = \frac{4}{5} \: }}}$

More to know :-

$\boxed{ \tt{ \: sinx = x - \frac{ {x}^{3} }{3!} + \frac{ {x}^{5} }{5!} + - - - }}$

$\boxed{ \tt{ \: cosx = 1 - \frac{ {x}^{2} }{2!} + \frac{ {x}^{4} }{4!} + - - - }}$

$\boxed{ \tt{ \: {e}^{x} = 1 + x + \frac{ {x}^{2} }{2!} + \frac{ {x}^{3} }{3!} + - - - }}$

$\boxed{ \tt{ \: {e}^{ - x} = 1 - x + \frac{ {x}^{2} }{2!} - \frac{ {x}^{3} }{3!} + - - - }}$