# `lim_(x->0) (x^2)/(1 - cos(x))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t...

`lim_(x->0) (x^2)/(1 - cos(x))` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

*print*Print*list*Cite

`lim_(x->0) x^2/(1-cos(x))`

The function `f(x) = x^2/(1-cosx)` is undefined at x=0. So to compute its limit as x

approaches zero, apply the L'Hospital's Rule. Take the derivative of the numerator and denominator.

`=lim_(x->0) ((x^2)')/((1-cos(x))')= lim_(x->0) (2x)/sin(x)`

The resulting function `(2x)/sin(x)` is still undefined at x=0. So, take the derivative of the numerator and denominator again.

`=lim_(x->0) ((2x)')/((sin(x))')= lim_(x->0) 2/cos(x) `

Now that the resulting function is defined at x=0, proceed to plug-in this value of x.

`= 2/cos(0) = 2/1=2`

**Therefore, `lim_(x->0)x^2/(1-cos(x))=2.`**