Question

Solve: $\underset{x\to 0}{lim}\frac{1-\cos x}{x^2}$

Answer 1

Consider the given limits.

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

We know that

$\cos x=1-2{\sin }^2\frac{x}{2}$

$2{\sin }^2\frac{x}{2}=1-\cos x$

Therefore,

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

${lim}_{x\to 0}\frac{2{\sin }^2\frac{x}{2}}{4\times {\left(\frac{x}{2}\right)}^2}$

We know that,

${lim}_{x\to 0}\frac{\sin x}{x}=1$

Therefore,

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

$=\frac{1}{2}$

Hence, the value of limit is $\frac{1}{2}$.