Question

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

Answer 1

We have $\cos x=1-2{\sin }^2\left(\frac{x}{2}\right)$ using sub-multiple angle formula.

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

$={lim}_{x\to 0}\frac{1-(1-2{\sin }^2\left(\frac{x}{2}\right))}{{\sin }^{2}x}$

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

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

$={lim}_{x\to 0}\frac{2\times 1\times \frac{x^2}{4}}{x^2}$ since ${lim}_{\theta \to 0}\frac{\sin \theta }{\theta }=1$

$=\frac{1}{2}$ by simplifying.