Prove that the function $$f(x)=e^x:=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$ is continuous at $x_0=1$ using the delta epsilon definition of continuous, which is:

$$\forall \varepsilon >0 \exists \delta>0 (\forall x\in D:|x-x_0|<\delta) |f(x)-f(x_0)|<\varepsilon$$

Since, in this particular context, the limit inside $e^x$ was proved to be convergent for all $x$ using the Bernoulli inequality and monotone convergence theorem, I'm struggling to see how I can apply a delta epsilon proof here. In my limited experience, I've applied it to nothing more than simple algebraic functions, but this is pretty significantly different. How do I start?

  • $\begingroup$ Hint: How would you convert a limit to an epsilon delta definition? $\endgroup$
    – MasterYoda
    Apr 2 '18 at 16:47
  • 2
    $\begingroup$ Shouldn't it be $$e^x := \lim_{n\to\infty}\left(1+\frac xn\right)^n\;\; \mathsf ? $$ $\endgroup$
    – Math1000
    Apr 2 '18 at 17:30
  • $\begingroup$ Yes, sorry, I have fixed it. $\endgroup$
    – jippyjoe4
    Apr 3 '18 at 0:12

You proved previously that the limit is convergent. This means $$\forall\epsilon>0\;\exists N\in\mathbb{N}\;{s.t.}\;n\geq N\implies\left|e^x-\left(1-\frac{x}{n}\right)^n\right|<\epsilon$$ To prove continuity of the defined function, we want to show $$\left|f(x)-f(x_0)\right|=\left|\lim_{n\to\infty}\left(1-\frac{x}{n}\right)^n - \lim_{n\to\infty}\left(1-\frac{x_0}{n}\right)^n\right|<\epsilon$$ Our approach will be to reduce this to $$\left|e^x-e^{x_0}\right|<\epsilon$$ which is a well known and easy to prove continuous function. Let's start by choosing $N$ so that the sequence is convergent. Now let's cleverly add zero. $$\begin{align}\left|\left(1-\frac{x}{n}\right)^n - e^x + e^x - \left(1-\frac{x_0}{n}\right)^n\right|&\leq\left|e^x-\left(1-\frac{x}{n}\right)^n\right|+\left|e^x-\left(1-\frac{x_0}{n}\right)^n\right|\\&<\frac{\epsilon}{3}+\left|e^x-\left(1-\frac{x_0}{n}\right)^n\right|\\ &=\frac{\epsilon}{3}+\left|e^x-e^{x_0}+e^{x_0}-\left(1-\frac{x_0}{n}\right)^n\right|\\ &<\frac{2\epsilon}{3}+\left|e^x-e^{x_0}\right|\end{align}$$ We want to choose a $\delta$ such that $$\left|e^x-e^{x_0}\right|\leq\frac{\epsilon}{3}$$ Consider the case that $x\geq x_0$, then consider $e^x-e^{x_0}=e^{x_0}\left(e^{x-x_0}-1\right)$. So we can choose a $\delta$ such that $$e^{x_0}\left(e^\delta-1\right)=\frac{\epsilon}{3}\implies\delta=\ln\left(1+e^{-x_0}\frac{\epsilon}{3}\right)$$ which is what we wanted to show. The same can be shown for $x_0>x$. Note that this $\delta$ is dependent on $x_0$ indicating point-wise continuity.

Edit: Fixed based on Jason DeVito's counterexample.

  • $\begingroup$ What do you mean by "Choose $N$ large enough so that the limit is convergent"? $\endgroup$ Apr 2 '18 at 22:37
  • $\begingroup$ @JasonDeVito I meant to say sequence is convergent. Thanks for that. I changed it accordingly. $\endgroup$
    – MasterYoda
    Apr 3 '18 at 0:03
  • $\begingroup$ There was a mistake in my original post; the limit should have been defined as follows: $e^x:=\lim_{n\to\infty}(1+\frac{x}{n})^n$ Although this doesn't change your answer drastically, you should be aware of it. $\endgroup$
    – jippyjoe4
    Apr 3 '18 at 0:18
  • $\begingroup$ @MasterYoda: I'm still not sure what you mean. The sequence is convergent or not, independent of whether we pick a value for $N$ or not. $\endgroup$ Apr 3 '18 at 0:33
  • $\begingroup$ @JasonDeVito Thanks. Changed again. Keep in mind $N$ must be chosen to satisfy the logic statement. $\endgroup$
    – MasterYoda
    Apr 3 '18 at 0:40

Here is some hints that will help you. It's just write it with more details.

Let $x_0\in\mathbb R$ and $\epsilon>0$ be arbitrary real numbers.

Notice that $$|e^x - e^{x_0}| = |e^x - (1+x/n)^n + (1+x/n)^n - (1+{x_0}/n)^n + (1+{x_0}/n)^n - e^{x_0}| \leq |e^x-(1+x/n)^n| + |(1+x/n)^n-(1+{x_0}/n)^n| + |(1+{x_0}/n)^n-e^{x_0}|$$ so a $\epsilon/3$-argument would be a nice idea.

Of course $|e^x-(1+x/n)^n|$ and $|(1+{x_0}/n)^n-e^{x_0}|$ get small enough ($<\epsilon/3$) as $n$ goes to $\infty$.

So it's sufficient to show that $|(1+x/n)^n-(1+{x_0}/n)^n|$ get small enough ($<\epsilon/3$) as $x$ goes to $x_0$. Well, here we can use the fact that for a fixed $n$ the function $z\mapsto(1+z/n)^n$ is continuous at $x$ (it is a polynomial).

Hope it helps.

  • $\begingroup$ I think that showing $|(1+x/n)^n - (1+x_0/n)^n|$ gets small as $x\rightarrow x_0$ needs more than just the fact that for fixed $n$, $z\mapsto (1+z/n)^n$ is continuous. For each $n$, by continuity of polynomials, there is a $\delta_n$ with the property that $|x-x_0|<\delta_n$ implies $|(1+x/n)^n - (1+x_0/n)^n|<\epsilon/3$, but it could be the case that $\delta_n\rightarrow 0$ as $n\rightarrow \infty$. (This is what happens, for example, if one uses $z\mapsto z^n$ instead). What one needs is a uniform $\delta$ which is independent of $n$. $\endgroup$ Apr 3 '18 at 0:41
  • $\begingroup$ We can choose $N\in\mathbb N$ so that both $|e^x-(1+x/N)^N|$ and $|e^{x_0}-(1+{x_0}/N)^N|$ are less than $\epsilon/3$ and then work with $z\mapsto(1+z/N)^N$. Am I missing something? $\endgroup$ Apr 3 '18 at 0:45
  • $\begingroup$ I may be missing something ;-). What if you apply your argument to $f(x):= \lim_{n\rightarrow \infty} x^n$, where I'm thinking of $x\in [0,1]$ For each $n$, $f_n(x) = x^n$ is continuous, but $f = \lim_{n\rightarrow \infty} f_n$ is the discontinuous function which is $0$ on $[0,1)$ and $1$ at $x=1$. $\endgroup$ Apr 3 '18 at 0:49
  • $\begingroup$ Ah - I got it: the issue is not where I said it was earlier. I agree (now) that once you find an appropriate $N$, that final inequality follows from continunity. But I'm not sure you can find the $N$ like you want. The problem is that for each $x$, there is an $N = N(x)$ which makes the term $|e^x - (1+x/n)^n|$ small, but as $x$ varies, this $N$ might blow up to infinity. $\endgroup$ Apr 3 '18 at 0:54
  • $\begingroup$ This, in turn, can cause the relevant $\delta$ in the last step to shrink to $0$. This is a problem because the definition of continuity at $x_0$ begins with "For all $\epsilon > 0$, there is a $\delta > 0$ with the property that for all $x$...." so $\delta$ must be independent of $x$. My analysis is rusty, so I don't know how to fix it in this particular problem, but the key is that $(1+x/n)^n\rightarrow e^x$ is not just point wise convergence, but uniform convergence. $\endgroup$ Apr 3 '18 at 1:00

Since you already covered the hard part of showing that the limit of sequence $(1+(x/n))^{n}$ exists for all real $x$, the continuity of the function defined via this limit will not seem that difficult to you.

Establish that $e^{x+y} =e^xe^y$ and then note that $$|e^x-e|=e|e^{x-1}-1|$$ and then use the inequality (you need to prove this too) $$1+x\leq e^x\leq \frac{1}{1-x}$$ for $0<x<1$. This should help you to prove that $e^x$ is continuous at $1$ by finding an appropriate $\delta$.

Let me know if you have difficulty proceeding in this manner and then I will provide more details. As a bonus that inequality mentioned above is fundamental and powerful and can be used to prove that $e^x$ is differentiable everywhere with the derivative equal to itself.


I am going to initially write $\exp(x)$ instead of $e^x$ to lesssen the tempatation to use exponential rules, until we show $\exp(x)$ really is an exponential function. (I am assuming you have already proved some exponential laws in your class).

Assuming $x > 0$, Set $m = n/x$, so that $m\rightarrow \infty$ as $n\rightarrow \infty$. Substituting this in gives $\exp(x):=\lim_{n\rightarrow \infty} (1 + x/n)^n = \lim_{m\rightarrow \infty} (1 + 1/m)^{mx} = [\lim_{m\rightarrow \infty} (1+1/m)^m]^x = \exp(1)^x$. Defining $e:=\exp(1)$, this proves $\exp(x) = e^x$, so it really is an exponential.

We next claim that if $e^x$ is continuous at $x_0 = 0$, then $e^x$ is continuous everywhere. To see this, choose $x_0\in\mathbb{R}$ suppose $\epsilon>0$ is given. Then $|e^x - e^{x_0}| = |e^{x_0}(e^{x-x_0} - 1)| = e^{x_0}|e^{x-x_0} - 1|$. By hypothesis, there is a $\delta$ with the property that if $|y|<\delta$, then $|e^y - 1| < \frac{\epsilon}{e^{x_0}}$. Then using this $\delta$, if $y=|x-x_0| < \delta$, then $|e^x - e^{x_0}| = e^{x_0}|e^{x-x_0} - 1| < e^{x_0}\frac{\epsilon}{e^{x_0}} = \epsilon$, as desired.

We next claim that if $e^x$ is continuous from the right at $x_0$, then it is automatically continuous from the left at $x_0$. The point is that $e^{x_0}\neq 0$, so $1/e^x $ is continuous from the right at $x_0$. But $1/e^x = e^{-x}$, so this is the same as being continuous from the left.

Thus, we need only show that $e^x$ is continuous from the right when $x_0=0$. So, let $\epsilon > 0$ be given. Now, pick $\delta < \min\{1, \frac{\epsilon}{\epsilon+1}\}$.

Assume $0<x<\delta$. Then using the binomial theorem, \begin{align*} |e^x - 1| &= |\lim_{n\rightarrow\infty} (1+x/n)^n -1| \\ &= \left|\lim_{n\rightarrow\infty} \sum_{k=0}^n {n\choose k}\left( \frac{x}{n} \right)^k -1\right|\\ &= \left| \lim_{n\rightarrow\infty}\sum_{k=1}^n {n\choose k}\frac{1}{n^k} x^k\right| \end{align*}.

We need to estimate ${n\choose k} \frac{1}{n^k}$. We get \begin{align*} {n\choose k}\frac{1}{n^k} = \frac{\overbrace{n(n-1)...(n-(k+1))}^{k\ terms}}{k!} \frac{1}{n^k} \\ &\leq \frac{n(n)...(n)}{k!} \frac{1}{n^k} \\ &= \frac{1}{k!}\\ &\leq 1\end{align*}

In particular, since $x> 0$, $$\left| \lim_{n\rightarrow\infty}\sum_{k=1}^n {n\choose k}\frac{1}{n^k} x^k\right|\leq \left|\sum_{k=1}^\infty x^k\right| = \left|\frac{x}{1-x}\right|,$$ where we use the geometric series formula (which is valid since $|x|<\delta < 1$.)

Now, one easily verifies that $\frac{x}{1-x} < \epsilon$ iff $x< \frac{\epsilon}{1+\epsilon}$, and since $|x|<\delta < \frac{\epsilon}{1+\epsilon}$. This completes the proof that $f(x) = e^x$ is continuous.


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