$\lim_{x \to -\infty}\left(1+ \frac{1}{x}\right)^x$

Question

I'm a bit rusty with limits but my book says that

$$ \lim_{x \to -\infty}\left(1+ \frac{1}{x}\right)^x= e$$ and I don't agree completely, knowing that $$ \lim_{x \to +\infty}\left(1+ \frac{1}{x}\right)^x=e$$

In my opinion the right result should be $1/e$ but I'm not completely sure.

Answer 1

Note that the $x$ in the exponent also goes to $-\infty:$

$$\lim\limits_{x \to -\infty} (1 + \frac 1 x ) ^ x = \lim\limits_{x \to \infty} (1 - \frac 1 x ) ^{-x} = \frac 1 {\lim\limits_{x \to \infty} (1 - \frac 1 x) ^ x} = \frac 1 { \frac 1 e} = e$$

Answer 2

(From Demidovich book) The statement is true. In general if $\lim_{x\to a}f (x)=1$ and $\lim_{x\to a}g(x)=\pm\infty$, then $$\lim_{x\to a}f (x)^{g (x)}=e^{\ \displaystyle{\lim_{x\to a}(\,f(x)-1)g(x)}} $$

Answer 3

The general formula is $$\lim_{x\to\infty}\Bigl(1+\frac ax\Bigr)^x=\mathrm e^a,$$ and its proof relies on the high-school limit: $$\lim_{u\to 0}\frac{\ln(1+au)}u=a,$$ which corresponds to the very definition of the derivative of the function $\ln(1+au)$.