# $\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$

$\lim_{x \to 0}\frac{\sin^2x-x^2}{x^2\sin^2x}$

I just can't do anything with this besides l'Hospital's rule (which doesn't seem to be a good idea). Can you help me, please?

• Are Taylor series out of the question too? – user170231 Nov 17 '15 at 16:53
• L'Hospital is not a good idea? Why not? – imranfat Nov 17 '15 at 16:54
• If you use $sin^2x=\frac{1}{2}-\frac{1}{2}cos2x$ and $cos^2x=\frac{1}{2}+\frac{1}{2}cos2x$ and use these identities in your limit, L'Hospital will go easier. – imranfat Nov 17 '15 at 16:58

Since $x \approx \sin x$ when $x \rightarrow 0$, the denominator is of order $x^4$.
The numerator needs more careful analysis of $\sin x$. Using $\sin x = x - \frac{x^3}{6} + o(x^3)$, we get $\sin^2 x = x^2 - \frac{x^4}{3} + o(x^4)$, so the numerator is $\sin^2 x - x^2 \approx -\frac{x^4}{3}$.
Dividing and ignoring terms of lower order, we get the limit: $-\frac{1}{3}$.