Originally Posted by

**Henryt999** Im sorry som error in the laTex script, anyway

The solution is telling me that the limit is not 1 but 0.

Here is the books "correct solution".

$\displaystyle \lim_{x \to 0} \frac{x^2}{1-cos^2x} = \lim_{x \to 0} \frac{x}{1-cosx} \times \frac {x}{1+cosx} $

and then since the $\displaystyle \frac{x}{1-cosx} \rightarrow \frac{0}{1}$ and

and the $\displaystyle \frac{x}{1+cosx} \rightarrow \frac{0}{1+1}$ then

$\displaystyle \lim_{x \to 0} = \frac{0}{1} \times \frac{0}{1+1} = 0$