Question

Evaluate
$\underset{x\to 0}{lim}\frac{x^2+1-\cos x}{x\sin x}$

Answer 1

$\underset{x\to 0}{lim}\frac{x^2+1-\cos x}{x\sin x}$

it is of the form $\frac{0}{0}$

So we will apply L hospital rule until the result gives a suitable result (determinate form)

$\underset{x\to 0}{lim}\frac{\frac{d}{dx}(x^2+1-\cos x)}{\frac{d}{dx}x\sin x}=\underset{x\to 0}{lim}\frac{2x+\sin x}{\sin x+x\cos x}$

Again Applying Same Rule by differentiating both numerator and denominator

$\underset{x\to 0}{lim}\frac{2+\cos x}{-x\sin x+\cos x+\cos x}=\frac{2+1}{0+2}=\frac{3}{2}$