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This question already has an answer here:

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.

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marked as duplicate by Xander Henderson, Did limits Apr 7 '18 at 19:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ 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 $0$? $\endgroup$ – Andrew Li Apr 7 '18 at 16:42
  • $\begingroup$ @AndrewLi no it is not legal $\endgroup$ – user Apr 7 '18 at 16:55
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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 $$ \lim_{x\to-\infty} \left( 1+\frac{1}{x} \right)^{x^2} =\lim_{u\to0-}\left( 1+u \right)^{1/u^2}=0. $$

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

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

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