How to find type of singularity and residue of $f(x)=\sin\left(e^{1/x}\right)$?

Question

How to find type of singularity and residue of $f(x)=\sin\left(e^{1/x}\right)$.

Singularity is $x_0=0$, but how do I find limit of this function to determine type of singularity?

Answer 1

Notice that as $x\to0^+$ we have $e^{1/x}\to+\infty$ so that in particular $f$ oscillates wildly in $[-1,1]$ as $x\to0^+$. Therefore, the singularity is not removable.

Moreover, because the real sine is bounded, the singularity cannot be a pole $(|f|$ remains bounded near $x_0=0$ for real values of $x)$. It follows that the only possibility is that $x_0$ is an essential singularity.

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EDIT: Generally, one uses residue when referring to meromorphic functions, that is, functions which are holomorphic except at some isolated points, which are poles. However, we may define residue more generally as follows:

For any pair of radii $0\leq r<R$, any holomorphic function $f$ defined on the annulus

$$A(r,R,z_0)=\{z\in\mathbb{C}\,|\,r<|z-z_0|<R\}$$

has a unique Laurent series expansion. This means there are unique coefficients ${(a_k)}_{k\in\mathbb{Z}}$ with

$$f(z)=\sum_{k=-\infty}^{+\infty}a_k\,{(z-z_0)}^k\tag{$*$}$$

for all $z \in A(r,R,z_0)$. The residue of $f$ at $z_0$ is the coefficient $a_{-1}$ when $r=0$ (notice that it does not depend on the value of $R$).

By uniqueness of the Laurent series expansion, however you find the coefficients of a series of the form $(*)$, the resulting series will be the Laurent expansion (provided, of course, it converges to $f$).

Can you think of any ways to find the Laurent series expansion for $f$ around $x_0=0$?

Answer 2

The question about identifying the singularity has been answered well in the other answer, so I will answer about calculating the residue: $$f(z)=\sin(e^{1/z})=e^{1/z}-\frac{e^{3/x}}{3!}+\frac{e^{5/x}}{5!}-...\\=(1+\frac1z+\frac1{2!z^2}+...)-\frac{1}{3!}(1+\frac3z+\frac3{2!z^2}+...)+\frac{1}{5!}(1+\frac5z+\frac5{2!z^2}+...)-...$$ So the residue, which is the coefficient of $z^{-1}$ is $$\frac1{0!}-\frac1{2!}+\frac1{4!}-\frac1{6!}+...=\cos(1)\approx0.54$$