Evaluate the limit $\lim_\limits{x\to-\infty} (1+\frac{1}{x})^{x^2}$

Question

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

I can substitute in $x := -x$ to show that this is the same as $$ \lim_{x\to\infty} \left(1-\frac{1}{x}\right)^{x^2}$$

However, I don't know where to proceed from here. I can rewrite the limit as

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

but this does not help as Algebra of Limits Product only applies to finitely many limits. I also tried substituting $x := 1/x$ to get

$$ \lim_{x\to0^+} (1-x)^{\frac{1}{x^2}} $$ but this does not help me either. I expect intuitively the answer to be $\infty$ as the larger exponent means the limit is "growing faster" than $\lim\limits_{x\to\infty} (1+\frac{1}{x})^{x} = e$ but I do not know how to show this.

Answer 1

Make the substitution $u=1/x$ so $$ \lim_{x\to-\infty} \left( 1+\frac{1}{x} \right)^{x^2} =\lim_{u\to0-}\left( 1+u \right)^{1/u^2} $$ Taking logarithms, note that

$$ \frac{1}{u^2}\log(1+u)=\frac{1}{u}\frac{\log(1+u)}{u}\to-\infty $$ as $u\to 0-$since $$ \log(1+u)/u\to 1 $$ by the definition of the derivative and $1/u\to-\infty$. Thus $$ \lim_{x\to-\infty} \left( 1+\frac{1}{x} \right)^{x^2} =\lim_{u\to0-}\left( 1+u \right)^{1/u^2}=0. $$

Answer 2

Hint: $(1-1/x)^x\to 1/e$ as $x\to\infty$, which specifically means that from some point on, we have $0<(1-1/x)^x<1/2$.

Answer 3

As an alternative to squeeze theorem, let $x=-y\to \infty$

$$\lim_{x\to-\infty} \left(1+\frac{1}{x}\right)^{x^2}=\lim_{y\to \infty} \left(1-\frac{1}{y}\right)^{y^2}$$

and

$$ \left(1-\frac{1}{y}\right) ^{y^2}=e^{y^2\log{\left(1-\frac{1}{y}\right)}}=e^{-y\,\frac{\log{\left(1-\frac{1}{y}\right)}}{-\frac1y}}\to 0$$

indeed by standard limits $\frac{\log{\left(1-\frac{1}{y}\right)}}{-\frac1y}\to 1$.