## Limit Problem 1

Compute $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{1-\cos x}$ ? 😀

Solution

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

We multiply the numerator and denominator by $1+\cos x$ and use the fact that $\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sin x}=1$ Therefore:

$\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}(1+\cos x)}{(1-\cos x)(1+\cos x)}=\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{x}{\sin x} \right)}^{2}}\underset{x\to 0}{\mathop{\lim }}\,(1+\cos x)=2$

**Another solution, using L’Hopital’s rule, is possible: $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2x}{\sin x}=2$

Thank you 😀