Question

Evaluate the following

$lim \: x \to \: 0 \: \frac{1 - cos2x}{cos2x - cos4x}$

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Answer 1

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

Given expression is

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \: \frac{1 - cos2x}{cos2x - cos4x} \: }$

If we substitute directly x = 0, we get

$\rm \:  =  \: \dfrac{1 - 1}{1 - 1}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form.

So, to evaluate

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \: \frac{1 - cos2x}{cos2x - cos4x} \: }$

We know

$\boxed{ \tt{ \: 1 - cos2x = {2sin}^{2}x \: }}$

$\boxed{ \tt{ \: cosx - cosy =2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{y - x}{2} \bigg] \: }}$

So, using this, we get

$\rm \:  =  \: \displaystyle\lim_{x \to 0} \: \frac{ {2sin}^{2} x}{2sin\bigg[\dfrac{2x + 4x}{2} \bigg]sin\bigg[\dfrac{4x - 2x}{2} \bigg]}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0} \: \frac{ {sin}^{2} x}{sin\bigg[\dfrac{ 6x}{2} \bigg]sin\bigg[\dfrac{2x}{2} \bigg]}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0} \frac{ {sin}^{2} x}{sin3x \: sinx}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0} \frac{sinx}{sin3x}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0} \frac{sinx}{x} \times x \times \displaystyle\lim_{x \to 0} \frac{3x}{sin3x} \times \dfrac{1}{3x}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0} \frac{sinx}{x} \times \displaystyle\lim_{x \to 0} \frac{3x}{sin3x} \times \dfrac{1}{3}$

We know,

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

So, using this, we get

$\rm \:  =  \: 1 \times 1 \times \dfrac{1}{3}$

$\rm \:  =  \: \dfrac{1}{3}$

Therefore,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \: \frac{1 - cos2x}{cos2x - cos4x} \: } = \frac{1}{3} \: }}$

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Additional Information :-

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

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

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

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

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