1
$\begingroup$

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

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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: $-\frac{1}{3}$.

$\endgroup$
0
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.