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

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?

  • $\begingroup$ Are Taylor series out of the question too? $\endgroup$ – user170231 Nov 17 '15 at 16:53
  • 3
    $\begingroup$ L'Hospital is not a good idea? Why not? $\endgroup$ – imranfat Nov 17 '15 at 16:54
  • $\begingroup$ 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. $\endgroup$ – 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: $-\frac{1}{3}$.


Hint: \begin{equation*} \frac{\sin ^{2}x-x^{2}}{x^{2}\sin ^{2}x}=\left( \frac{\sin x-x}{x^{3}}% \right) \left( \frac{\sin x+x}{x}\right) \left( \frac{x}{\sin x}\right) ^{2} \end{equation*} \begin{eqnarray*} \lim_{x\rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right) &=&-\frac{1}{6} \\ \lim_{x\rightarrow 0}\left( \frac{\sin x+x}{x}\right) &=&1+1=2 \\ \lim_{x\rightarrow 0}\left( \frac{x}{\sin x}\right) ^{2} &=&1^{2}=1. \end{eqnarray*}


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