Question

Find the value of $L=\underset{x\to 0}{lim}\frac{\sec x-1}{x^2{\left(\sec x+1\right)}^2}$

Answer 1

${lim}_{x\to 0}\frac{\sec x-1}{x^2{(\sec x+1)}^2}$

$={lim}_{x\to 0}\frac{\frac{1}{\cos x}-1}{x^2{(\frac{1}{\cos x}+1)}^2}$

$={lim}_{x\to 0}\frac{1-\cos x}{\cos x\times \frac{x^2{(1+\cos x)}^2}{{\cos }^{2}x}}$

$={lim}_{x\to 0}\frac{2{\sin }^2\frac{x}{2}\cos x}{x^2{\left(2{\cos }^2\frac{x}{2}\right)}^2}$

$={lim}_{x\to 0}\frac{2{\sin }^2\frac{x}{2}}{\frac{x^2}{2}\times 2}\times \frac{\cos x}{4{\cos }^4\frac{x}{2}}$

$=\frac{2}{2}\times \frac{1}{4}=\frac{1}{4}$