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Find $\lim_{z\rightarrow 0}\left(\frac{1}{z^{2}}-\frac{1}{\sin^{2}z}\right)$. First I let $z=x+iy$ and then substitute to the limit and I let $y=0$ so it now limit $x\rightarrow 0$. So i use L'Hopital but I didn't get the answer.

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  • $\begingroup$ L'Hopital's rule wouldn't work here, because the limit is of the form $\infty - \infty$. Have you tried finding a common denominator and writing the expression as one fraction, and from there perhaps applying l'Hopital or using Taylor series? $\endgroup$ – Stahl Feb 26 '16 at 14:12
  • $\begingroup$ Write it as $\dfrac{\sin^2 z - z^2}{z^2\sin^2z}$, and Taylor-expand the squared sine in the numerator a bit to get the limit. $\endgroup$ – Daniel Fischer Feb 26 '16 at 14:15
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Use Taylor series: when $z \approx 0$, $\sin z \approx z - {z^3 \over 6}$, so $\sin^2 z \approx z^2 - \frac{z^4}{3}$

Now,

$$\frac{1}{z^2} - \frac{1}{\sin^2{z}} \approx \frac{1}{z^2}\left(1 - \frac{1}{1 - \frac{z^2}{3}}\right) = \frac{1}{z^2}\frac{-\frac{z^2}{3}}{1-\frac{z^2}{3}} = -\frac{1}{3}\frac{1}{1-\frac{z^2}{3}} \rightarrow -\frac{1}{3}$$

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$$\frac{1}{z^2}-\frac{1}{\sin^2 z} = \frac{(\sin z-z)\cdot(\sin z+z)}{z^2\sin z\cdot\sin z}$$ but since $\lim_{z\to 0}\frac{\sin z}{z}=1$ we have: $$ \frac{1}{2}\lim_{z\to 0}\left(\frac{1}{z^2}-\frac{1}{\sin^2 z}\right) = \lim_{z\to 0}\frac{\sin z-z}{z^2\sin z} = \lim_{z\to 0}\frac{\sin z-z}{z^3} \stackrel{DH}{=}\lim_{z\to 0}\frac{\cos z-1}{3z^2}=-\frac{1}{6}$$ since $1-\cos(z) = 2\sin^2\frac{z}{2}$. Our limit is so $\color{red}{-\frac{1}{3}}$.

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