Evaluating the limit $\lim_{x \to -\infty} \left(1+ \frac{1}{x}\right)^{x^2}$

Question

I am trying to evaluate $$ \lim_{x \to -\infty} \left(1+ \frac{1}{x}\right)^{x^2}. $$ I'd say it tends to 0, 1 or something linked to $e$ but I have no clue how to prove this... I'm getting really stuck as I try to clarify things. Can you hint me? Thank you.

Answer 1

Since $(1+1/x)^x \to e$, this $\to \infty$.

Added: I read the limit as $\infty$, not $-\infty$. In this case, if $y=-x$ so $y \to +\infty$, $(1+1/x)^x = 1/(1-1/y)^y \to 1/(1/e) =e$, so this $\to e^x =e^(-y) =1/e^y \to 0$.

Answer 2

You can use substitution:

$$y=-x\implies x\to-\infty\iff y\to\infty\;,\;\;\text{and thus}:$$

$$\lim_{x\to-\infty}\left(1+\frac1x\right)^{x^2}=\lim_{y\to\infty}\left[\left(1-\frac1y\right)^y\right]^y=0$$

since

$$\lim_{y\to\infty}\left(1+\frac ay\right)^y=e^a\;\;\forall\,a\in\Bbb R$$