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How to find this limit (if it exists) : $\lim_{z \to 0} \left({\dfrac{\sin(z)}{z}}\right)^{\frac{1}{z^2}}$ where $z$ is a complex number.

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    Try finding it. You might as well able to do that. – 2017-02-14
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    I did try finding it but found that it equals to one. I used this identity : x^a = e^(a*ln(x)) then did a taylor expansion for ln(sin(z)/z) simplified and found one is that correct ? – 2017-02-14
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    From now onward please try to write a little bit of what you did with the question. – 2017-02-14

4 Answers 4

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Let $y=\left({\dfrac{\sin z}{z}}\right)^{\frac{1}{z^2}}$ then $\ln y=\dfrac{\ln\sin z-\ln z}{z^2}$ two times L'Hospital rule shows $$\lim_{z \to 0} \ln y=\lim_{z \to 0} \frac{-\sin z}{4\sin z+2z\cos z}=-\frac16$$ so $\lim_{z \to 0} y=e^{-\frac16}$

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For small $z$, $\frac{\sin z}{z}\approx 1-\frac{z^2}{6}\approx \exp -\frac{z^2}{6}$ so the limit is $\exp -\frac{1}{6}$.

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    how can 1 be close to exp ? – 2017-02-14
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    $1+x\approx e^x$ for small $x$. – 2017-02-14
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    ah it's exp(-z^2/6) thought it was exp(1) - z^2/6 was confused for a moment. thank you for shedding light. – 2017-02-14
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Write it as $$(1+z^2f(z))^{\frac{1}{z^2}}$$

where $$f(z)=\frac{\sin z-z}{z^3}$$

Thus you get the limit $$e^{-\frac{1}{6}}$$

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Hint:

Rewrite the limit as $$\lim_{z \to 0} \left({\dfrac{\sin z}{z}}\right)^{\frac{1}{z^2}} \stackrel{!}{=} \exp \lim_{z\to 0} \frac{1}{z^2}\ln \left(\frac{\sin z}{z} \right)$$ and apply L'Hospital.