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Find $\displaystyle\lim_{z\to0} \operatorname{pv}\left(\cos(z)^\frac{1}{z^2}\right)$.

Pv stands for principal value. I believe there is a certain trick I am missing. I am applying the fundamental definitions of the principal value of log, but this produces ugly calculations. Thank you all in advanced for your help.

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Recalling that $\left(1+\dfrac an\right)^n\to e^a$ as $n\to\infty$, it follows that $\left(1-\dfrac{z^2}{2}\right)^{1/z^2}\to e^{-1/2}$ as $1/z^2\to\infty$. And $\cos z = 1-\dfrac{z^2}{2}+\text{higher-degree terms}$. – Michael Hardy Oct 10 '13 at 1:16
Excellent. Indeed there was a feisty trick. Thank you. – Gustavo Montano Oct 10 '13 at 5:44
@eXtremity : Thank you. I've now made my comment into a posted answer. – Michael Hardy Oct 10 '13 at 19:49
up vote 1 down vote accepted

I didn't post this as an answer before since it was somewhat less than complete, but the comments suggest maybe this is what was needed:

Recalling that $\left(1+\dfrac an\right)^n\to e^a$, it follows that $$ \left(1-\dfrac{z^2}{2}\right)^{1/z^2}\to e^{-1/2}\text{ as } \frac{1}{z^2}\to\infty. $$ And $$ \cos z = 1 - \frac{z^2}{2} + \text{higher-degree terms}. $$

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Which Michael Hardy's answer is excellent, you could also do it with L'H after taking logarithm: $$ \lim_{z\to 0}\frac{\log \cos z}{z^2} = \lim_{z\to 0}\frac{ - \sin z}{2z \cos z } = -\frac12 $$ because $\frac{\sin z}{z}\to 1$ and $\cos z\to 1$.

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Thank you for your answer. A few questions: 1) I see you have used that the limit of two functions f and g is equal to the product of the limit of g and the limit of f. This particular limit calculation you have done - is it plausible in the complex field? Second, you have "put" the limit inside the exponential function. Is this also valid in the complex field? I understand that in the real field, as long as the function is continuous, than such a process is good. @user98130 – Gustavo Montano Oct 10 '13 at 5:42

$$ \cos\left(z\right)^{1/z^{2}} = \exp\left(\ln\left(\cos\left(z\right)\right) \over z^{2}\right) \overbrace{\quad% \sim \exp\left(\ln\left(1 - z^{2}/2\right) \over z^{2}\right) \sim\quad} ^{\mbox{when}\ z\ \sim\ 0} \exp\left(-z^{2}/2 \over z^{2}\right) = \color{#ff0000}{\Large{\rm e}^{-1/2}} $$

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