Question

Evaluate the limit: $\underset{x\to 0}{lim}\frac{1-cos2x}{cos2x-cos8x}$

Answer 1

We have,

$\underset{x\to 0}{lim}\frac{1-cos2x}{cos2x-cos8x}$

It becomes 0/0 form.

$\because cosA-cosB=2sin\frac{A+B}{2}sin\frac{B-A}{2}$

$\because 1-cos2x=2si{n}^{2}x$

$\Rightarrow \underset{x\to 0}{lim}\frac{2si{n}^{2}x}{2sin\frac{2x+8x}{2}sin\frac{8x-2x}{2}}$

$=\underset{x\to 0}{lim}\frac{2si{n}^{2}x}{2sin5xsin3x}$

$=\underset{x\to 0}{lim}\frac{sinx\times sinx}{sin5xsin3x}$

On dividing numerator and denominator by $x^2$,

we get

$=\underset{x\to 0}{lim}\frac{\frac{sinx}{x}\times \frac{sinx}{x}}{\frac{sin5x}{5x}\times \frac{sin3x}{3x}\times 15}$

$=\frac{1\times 1}{1\times 1\times 15}[\because \underset{x\to 0}{lim}\frac{sinx}{x}=1]$


$=\frac{1}{15}$

Therefore,

$\underset{x\to 0}{lim}\frac{1-cos2x}{cos2x-cos8x}=\frac{1}{15}$