Question

Using L' Hospital rule, evaluate: $\underset{x\to 0}{lim}(\frac{1}{x^2}-\frac{\cot x}{x})$

Answer 1

$\underset{x\to 0}{lim}(\frac{1}{x^2}-\frac{\cot x}{x})$

$=\underset{x\to 0}{lim}\frac{\sin x-x\cos x}{x^2\sin x}$
$=\underset{x\to 0}{lim}\frac{\sin x-x\cos x}{x^3}.\underset{x\to 0}{lim}\frac{x}{\sin x}$
$=\underset{x\to 0}{lim}\frac{\sin x-x\cos x}{x^3}.1[\because \underset{x\to 0}{lim}\frac{x}{\sin x}=1]$
$=\underset{x\to 0}{lim}\frac{\cos x-(\cos x-x\sin x)}{3x^2}$
[$\frac{0}{0}$ form, applying L' Hospital's rule]
$=\underset{x\to 0}{lim}\frac{x\sin x}{3x^2}$
$=\underset{x\to 0}{lim}\frac{\sin x}{3x^{}}$
[$\frac{0}{0}$ form, applying L' Hospital's rule]
$=\underset{x\to 0}{lim}\frac{\cos x}{3}=\frac{1}{3}$