TITLE: Finding the limit of a function and its reciprocal
QUESTION [0 upvotes]: I would like to compute
$$\displaystyle\lim_{z \to 0} f(z) = \displaystyle\lim_{z \to 0} \frac{z \cos z - \sin z}{\frac{1}{3}\cos^3z-\cos z + \frac{2}{3}}$$
and also
$$\displaystyle\lim_{z \to 0} \frac{1}{f(z)} = \frac{\frac{1}{3}\cos^3z-\cos z + \frac{2}{3}}{z \cos z - \sin z}$$
I am struggling to calculate either of these. Is it as simple as recursively applying L'Hospital's rule since we obtain indeterminate forms in both?
I've also tried multiplying each by $\frac{ \cos z}{ \cos z}$ and $\frac{ \sin z}{ \sin z}$, but end up with the same indeterminate form every time. Any thoughts on computing these limits? I have graphed both of these and I know that $z=0$ is a pole of $f$ since $f(z) \to \infty$ as $z \to 0$and $\frac{1}{f(z)} \to 0$ as $z \to 0$. But I am struggling to show each of these things.

REPLY [2 votes]: We have $$ \cos^3z-3\cos z +2=\cos^3z-\cos z -2\cos z+2=(1-\cos z)(\sin^2z +2)$$ Therefore $$f(z):={z\cos z -\sin z\over {1\over 3}\cos^3-\cos z+{2\over 3}} ={3\over \sin^2z+2} \, {z\cos z -\sin z\over 1-\cos z}$$
Next
$${z\cos z -\sin z\over 1-\cos z}=-z+{z-\sin z\over 1-\cos z}$$
Hence the limit in question is equal $${3\over 2}\, \lim_{z\to 0}{z-\sin z\over 1-\cos z}$$ We have
$${z-\sin z\over 1-\cos z}={z-\sin z\over z^3}{z^2\over 1-\cos z}\,z$$
As
$$z-\sin z={z^3\over 6} +o(z^3),\qquad 1-\cos z={z^2\over 2}+o(z^3),\quad{\rm when}\ z\to 0$$ we get
$$\lim_{z\to 0}{z-\sin z\over z^3}{z^2\over 1-\cos z}\,z={1\over 6}\cdot 2\cdot 0=0$$
Summarizing $\lim_{z\to 0}f(z)=0$ and the limit of ${1\over f(z)}$ at $z=0$ does not exist.
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