Evaluation of this line integral $\int z^2\cdot e^\left(\frac{1}{z}\right)\cdot\sin\left(\frac{1}{z}\right)dz$

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where the contour is $$|z|=1$$

if I use residual theorem $$res f(z) = \lim\limits_{z\to 0}  z^3\cdot e^\left(\frac{1}{z}\right)\cdot \sin\left(\frac{1}{z}\right)$$

this limit is undefined .However , if i expand the function in its laurent series i can find coefficient of first negative power which is equal to $\left(\frac{2}{3}\right)$ How is this possible $?$

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The "quick" formulae for residue only works for simple poles and poles "of finite order". They won't work for essential singularities though. However, Laurent series expansion always works.