Show that the limit $\lim_{z\to 0} e^{-1/z}$ does not exist
Along real axis putting $z=x+i0$,
we have $\lim_{z\to 0} e^{-1/z}=\lim_{x\to 0} e^{-1/x}=-\infty$
Along imaginary axis putting $z=0+iy$,
we have $\lim_{z\to 0} e^{-1/z}=\lim_{y\to 0} e^{-1/iy}=\lim_{y\to 0} e^{i/y}=\infty$
I.e two different approach to $z=0$ give two different limit, so limit does not exist.
Is the proof correct?
Second evaluation of limit is correct? ($\lim_{y\to 0} e^{-1/iy}=\lim_{y\to 0} e^{i/y}=\infty$ )
Is there any other method to solve?
Best Answer
It suffices to consider z real. $$lim_{x \to 0^+} e^{- \frac{1}{x}}=0$$ $$lim_{x \to 0^-} e^{- \frac{1}{x}}= \infty$$ So the limit does not exist.
I should also probably point out a couple of errors in your proof: 1. When you take the real limit, you should notice that x approaching from the negative and positive directions give different answers, so saying $lim_{x \to 0} e^{- \frac{1}{x}}= \infty$ isn't correct. 2. When you take the imaginary limit, the answer is not $\infty$. $y$ is real so as $y \to 0$ all that's happening is $e^{i/y}$ is spinning round the unit circle in the complex plane, still not converging to anything, but also not tending to infinity.