Question

Evaluate the limit: $\underset{x\to 0}{lim}\frac{sec5x-sec3x}{sec3x-secx}$

Answer 1

We have,

$\underset{x\to 0}{lim}\left[\frac{sec5x-sec3x}{sec3x-secx}\right]$

It becomes 0/0 form.

$=\underset{x\to 0}{lim}\left[\frac{\frac{1}{cos5x}-\frac{1}{cos3x}}{\frac{1}{cos3x}-\frac{1}{cosx}}\right]$

$=\underset{x\to 0}{lim}\left[\frac{\left\{\frac{cos3x-cos5x}{cos3xcos5x}\right\}}{\left\{\frac{cosx-cos3x}{cosxcos3x}\right\}}\right]$

$=\underset{x\to 0}{lim}\left[\frac{cosx(cos3x-cos5x)}{cos5x(cosx-cos3x)}\right]$

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

$=\underset{x\to 0}{lim}\left[\frac{cosx\left\{2sin\left(\frac{3x+5x}{2}\right)sin\left(\frac{5x-3x}{2}\right)\right\}}{cos5x\left\{2sin\left(\frac{x+3x}{2}\right)sin\left(\frac{3x-x}{2}\right)\right\}}\right]$

$=\underset{x\to 0}{lim}\left[\frac{sin4x\times sinx\times cosx}{cos5x\times sin2x\times sinx}\right]$

$=\underset{x\to 0}{lim}[\frac{sin4x}{4x}\times \frac{4x}{\frac{sin2x}{2x}\times 2x}\times \frac{cosx}{cos5x}]$

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

$=[1\times \frac{2}{1}\times \frac{cos0}{cos(5\times 0)}]$

$=1\times 2\times 1$

$=2$