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

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.

## marked as duplicate by Xander Henderson, Did limits StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 7 '18 at 19:08

• Don't know if this is legal, but can't you write it as $\left(1-x/x^2\right)^{x^2}$ with $x\to\infty$ and get $e^{-x}$ as $x\to\infty$ which is $0$? – Andrew Li Apr 7 '18 at 16:42
• @AndrewLi no it is not legal – user Apr 7 '18 at 16:55

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 $$\lim_{x\to-\infty} \left( 1+\frac{1}{x} \right)^{x^2} =\lim_{u\to0-}\left( 1+u \right)^{1/u^2}=0.$$

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

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