Question

${lim}_{x\to 0}\frac{sec5x-sec3x}{sec3x-secx}$

Answer 1

${lim}_{x\to 0}\frac{sec5x-sec3x}{sec3x-secx}={lim}_{x\to 0}\frac{\frac{1}{cos5x}-\frac{1}{cos3x}}{\frac{1}{cos3x}-\frac{1}{cosx}}$

$={lim}_{x\to 0}\left(\frac{\frac{cos3x-cos5x}{cos3xcos5x}}{\frac{cosx-cos3x}{cosxcos3x}}\right)={lim}_{x\to 0}(\frac{cos3x-cos5x}{cosx-cos3x}\times \frac{cosxcos3x}{cos3xcos5x})$

[Applying Componendo and dividendo formula]

$={lim}_{x\to 0}(\frac{-2sin4xsin(-x)}{-2sin\left(2x\right)sin(-x)}\times \frac{cosx}{cos5x})={lim}_{x\to 0}(\frac{sin4x}{sin2x}\times \frac{cosx}{cos5x})$

$=\frac{{lim}_{x\to 0}sin4x\times {lim}_{x\to 0}cosx}{{lim}_{x\to 0}sin2x\times {lim}_{x\to 0}cos5x}=\left(\frac{{lim}_{4x\to 0}\frac{sin4x}{4x}\times 4x}{{lim}_{2x\to 0}\frac{sin2x}{2x}\times 2x}\right)\times \left(\frac{{lim}_{x\to 0}cosx}{{lim}_{x\to 0}cos5x}\right)$

$=\frac{(1\times 4x)\times 1}{1\times 2x\times 1}[\because {lim}_{x\to 0}\frac{sinx}{x}=1\text{and}{lim}_{x\to 0}cosx=cos0=1]$

$=\frac{4x}{2x}=2$