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I’m generally pretty good at proving limits using epsilon-delta, but on this one I’m stuck — and have been for days. This is the problem I’m talking about:

\begin{equation} \lim_{x\to\infty}\frac{e^x}{e^x + x} = 1. \end{equation}

I know that I need to let $\epsilon>0$ and choose an $N$ such that if $x>N$, then $\big|\big(\frac{e^x}{e^x + x}\big) - 1\big| < \epsilon$. I can’t quite see how I might go about doing that though.

I would very much appreciate hints rather than complete solutions. I feel like I should be able to do this, but I definitely need a subtle push in the right direction. This might be embarrassingly simple, I just can’t spot how to go about it.


If anyone is curious about how a final answer might look, I guess I should put it up here. It takes as a given that $e^x > x^2$, because this is something that is not examinable in the class I’m taking.

Question: Prove carefully, using the definition of a limit, that

\begin{equation*} \lim_{x\to\infty}\frac{e^x}{e^x + x} = 1. \end{equation*}

Let $\epsilon>0$.

Let $N = \frac{1}{\epsilon}$.

Suppose that $x>N$.

\begin{equation*} \therefore x > \frac{1}{\epsilon} > 0. \end{equation*}

\begin{equation*} \therefore 0 < \frac{1}{x} < \epsilon. \end{equation*}

Note that $e^{x}>x^{2}$.

\begin{equation*} \therefore 0 < \frac{x}{e^x + x} < \frac{x}{e^x} < \frac{1}{x} < \epsilon. \end{equation*}

\begin{equation*} \therefore \bigg|\frac{x}{e^x + x}\bigg| < \epsilon. \end{equation*}

\begin{equation*} \therefore \bigg|\frac{-x}{e^x + x}\bigg| < \epsilon. \end{equation*}

\begin{equation*} \therefore \bigg|\frac{e^x - (e^x + x)}{e^x + x}\bigg| < \epsilon. \end{equation*}

\begin{equation*} \therefore \bigg|\frac{e^x}{e^x + x} - 1\bigg| < \epsilon. \end{equation*}

Therefore,

\begin{equation*} \lim_{x\to\infty}\frac{e^x}{e^x + x} = 1. \end{equation*}

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  • $\begingroup$ You could use, for example, L'Hopital's rule. $\endgroup$ – Michael Dec 10 '15 at 2:03
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    $\begingroup$ Or note that $(\frac{e^x}{e^x+x})- 1 = \frac{-x}{e^x+x}$. Or use $\frac{e^x}{e^x+x} = \frac{1}{1 + xe^{-x}}$. Can you prove something about $xe^{-x}$? $\endgroup$ – Michael Dec 10 '15 at 2:04
  • $\begingroup$ Are you allowed to use so-called "sandwich theorem"? $\endgroup$ – user160738 Dec 10 '15 at 2:13
  • $\begingroup$ Unfortunately not. I’m not allowed any shortcuts. $\endgroup$ – Henrik Dec 10 '15 at 2:22
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Hint. Rewrite as $$|\frac{x}{e^x+x}|$$

Note that:

$$\frac{x}{e^x+x}<\frac{x}{e^x}<\frac{1}{x}$$

for positive $x$.

The last inequality is nontrivial. You can prove this using the definition of $e^x$: $e^x \colon = \lim\limits_{n\to\infty} \left(1+\frac{x}{n}\right)^n>x^2$

Note that this is an increasing sequence, and use binomial theorem.

Or the other definition of $e^x$:

$$e^x \colon = 1+x+x^2+\dots>x^2$$

Let me know if you need more of a hint.

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  • $\begingroup$ That solved it for me, thanks! We haven’t learned the defintion of $e^{x}$, so knowing that it’s greater than $x^{2}$ really made the difference. $\endgroup$ – Henrik Dec 10 '15 at 4:29

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