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I'm having some trouble evaluating the $\lim_{x\to\infty}\frac{\left(1+\frac{1}{x}\right)^{x}}{x}$ in the context of real analysis. Denote this sequence by $(a_k)$

If so, then I need to prove that for all $\varepsilon>0$

$$\left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}-0\right| <\varepsilon$$

We know that $\lim_{x\to\infty}1/x=0$, however $|1/x|\leq\left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}\right|$, so I'm not sure this information is useful. I considered the sequential characterization of continuity, so if the sequence above converges to 0, then $f((a_k))$ converges to $f(0)$, but there is also not an obvious choice of $f$. Any suggestions into how to evaluate the limit?

amWhy
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jeb2
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    The numerator tends to $e$ (but if you aren't allowed to use this, then for this question it suffices to show that the numerator is bounded by some constant). – angryavian Mar 23 '21 at 06:56
  • @angryavian I noticed that about the numerator, but this is part of a proof of $e$ so I suppose I cannot use this fact. – jeb2 Mar 23 '21 at 06:58
  • Your limit equal to $e/ \infty=0$ Are you looking for $\varepsilon -\delta$ proof ? – lone student Mar 23 '21 at 07:02
  • @lonestudent yes Im trying come up with a $\varepsilon-\delta$ proof. – jeb2 Mar 23 '21 at 07:06

4 Answers4

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We will show that $$\lim_{x\to\infty}\frac{\left(1+\frac{1}{x}\right)^{x}}{x}=0.$$ We let $\varepsilon >0$ be given. We need to find a $\delta >0$ such that $|x-0|<\delta$ implies ${\displaystyle \left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}-0\right|<\varepsilon }$.

In this post, a user proved that $\left(1+\frac{1}{x}\right)^{x}<3.$ Also for every $x>1$, $1/x<x$. Choose $\delta=\varepsilon/3$. $${\displaystyle \left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}-0\right|<\frac{3}{x}<3x<3\delta<\varepsilon }.$$

vitamin d
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Since $\left(1+\frac{1}x\right)^x$ converges to $e$, for any constant $\epsilon > 0$ there exists $M>0$ such that for all $x>M$, $$0 \leq |a_k| \leq \frac{e+\epsilon}{|x|}.$$ Then $a_k \to 0$ from sandwich lemma.

Ganapathy K
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$\lim_{x \rightarrow \infty} (1+1/x)^x=e.$

Hence $x$ it is bounded above, say $M >0$ is a bound.

Then

$0<|(1/x)(1+1/x)^x| <|(1/x)M|;$

Take the limit $x \rightarrow \infty$ (Squeeze Theorem).

Peter Szilas
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$$y=\frac{\left(1+\frac{1}{x}\right)^x}{x}\implies \log(y)=x \log \left(1+\frac{1}{x}\right)-\log(x)$$ So, by Taylor $$\log(y)=x\Bigg[\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3 x^3}+O\left(\frac{1}{x^4}\right)\Bigg]-\log(x)$$ $$\log(y)=1-\log(x)-\frac{1}{2 x}+\frac{1}{3 x^2}+O\left(\frac{1}{x^3}\right)$$ $$y=e^{\log(x)}=\frac{e}{x}\left(1-\frac{1}{2 x}+\frac{11}{24 x^2}+O\left(\frac{1}{x^3}\right) \right)\sim \frac{e}{x} \quad \to 0$$

Claude Leibovici
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