# Determine the limit as x approaches 0 of (x^2)/(1 - cos x).

## Question:

Determine {eq}\; \lim_{x\rightarrow 0} \frac{x^2}{1 \, - \, \cos x} {/eq}.

## L' Hopital's Rule:

Suppose that we have to determine the following limit

{eq}\lim_{x\rightarrow a}\frac{f(x)}{g(x)} {/eq}

where either both {eq}f,\, and\, g {/eq} converge to zero or infinity. Then L' Hopital's rule implies that

{eq}\lim_{x\rightarrow a}\frac{f(x)}{g(x)} = \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}. {/eq}

For the problem

{eq}\lim_{x\rightarrow 0} \frac{x^2}{1 \, - \, \cos x} {/eq}

Notice that {eq}x^2 \rightarrow 0 {/eq} and {eq}1 - \cos x\rightarrow 0 {/eq} as {eq}x\rightarrow 0. {/eq}

Using L'Hopital's rule we have

{eq}\lim_{x\rightarrow 0} \frac{x^2}{1 \, - \, \cos x} = \lim_{x\rightarrow 0} \frac{2x}{\sin x} {/eq}

Again, {eq}2x \rightarrow 0 {/eq} and {eq}\sin x \rightarrow 0 {/eq} as {eq}x\rightarrow 0 {/eq}, so we can again use the aforementioned rule. Thus

{eq}[{MathJax fullWidth='false' \lim_{x\rightarrow 0} \frac{x^2}{1 \, - \, \cos x} = \lim_{x\rightarrow 0} \frac{2x}{\sin x} = \lim_{x\rightarrow 0 } \frac{2}{\cos x} = 2. }] {/eq}