Question

simply the following limit algebraically

limit as x-> 0 (x^2)/(1-cos^2(x))

Answer 1

\[1-\cos^2x=\sin^2x\]
So your limit is reduced to:
\[\lim_{x \to 0} \frac{x^2}{\sin^2x}=\lim_{x \to 0} \left ( \frac{x}{\sin x}\right )^2=\left ( \lim_{x \to 0} \frac{x}{\sin x}\right )^2=1\]
because the squaring function is continuous.