Does e = limit as x tends to negative infinity hold true?

Question

Does $$e=\displaystyle\lim_{x \to -\infty}\left(1+\frac{1}{x}\right)^x\qquad\quad?$$

Answer 1

$$\displaystyle\lim_{x \to -\infty}\left(1+\frac{1}{x}\right)^x$$ Put $x=-y$ $$=\displaystyle\lim_{y \to \infty}\left(1-\frac{1}{y}\right)^{-y}$$ $$=\lim_{y \to \infty} \frac{1}{\left(1-\frac{1}{y}\right)^y}=\frac{1}{e^{-1}}=e$$

Last step is from $$e^r=\displaystyle\lim_{y \to \infty}\left(1+\frac{r}{y}\right)^{y}$$

Answer 2

...or, in other words, rewrite the expression as $$ \lim_{y \to \infty} \frac{1}{\bigg(1-\frac{1}{y}\bigg)^y} $$

Answer 3

if base tends to 1 after taking its separate limit and power tends to infinity then use below formula this type of questions ...

       lim         (base-1)(power)
      x->infinity
    e