Proof verification: show that $\lim_{x\to 1}\frac{1}{e^{x-1}}$ does not exist.

Question

Show that the following limit does not exist: $$ \lim_{x\to 1}\frac{1}{e^{x-1}} $$

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Here is my attempt. Am I doing it correctly?

$$\lim_{x\to 1^+}\frac{1}{e^{x-1}} = + \infty,\quad\lim_{x\to 1^-}\frac{1}{e^{x-1}} = - \infty$$

Therefore $\lim_{x\to 1}\frac{1}{e^{x-1}}$ does not exist.

Answer 1

Note that

$$\lim_{x\to 1}\frac{1}{e^{x-1}}=\frac1{e^0}=1 $$

Answer 2

There is no singularity, just plug in: $\lim_{x\rightarrow 1}\dfrac{1}{e^{x-1}}=\dfrac{1}{e^{1-1}}=\dfrac{1}{e^{0}}=1$.

Answer 3

The limit exists and its value is 1. Probably a misprint.