# How do you prove lim h->0 of (1+h)^(1/h) exists?

I am planning to teach calculus to a student, and I would like to define Euler's number, $e$, as $\lim \limits_{h\to0} (1+h)^{\frac1h} .$ However, I do not know how to show that this limit exists (and is finite). Any help is appreciated!

Note by James Wilson
8 months, 2 weeks ago

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Start by proving that $x - \tfrac12x^2 < \ln(1+x) < x$ for all $x > 0$.

• If $f(x) = x - \ln(1+x)$, then $f'(x) = \tfrac{x}{1+x} > 0$ for all $x > 0$ and $f(0) =0$. The MVT gives $f(x) >0$ for $x > 0$.
• If $g(x) = \ln(1+x)-x+\tfrac12x^2$, then $g'(x) = \tfrac{x^2}{1+x} > 0$ for all $x > 0$, and $g(0)= 0$. The MVT gives $g(x) > 0$ for $x > 0$.

Then $-\tfrac12x < x^{-1}\ln(1+x) - 1 < 0$ for $x > 0$, and hence $\lim_{x \to 0+}x^{-1}\ln(1+x) = 1$. Given that exponentiation is continuous, we get the result.

However, we need to know about exponentiation and logarithms, and this comes a little after the existence of $e$, unless you take the approach below...

Define $e^x$ by its power series $E(x) \; = \; \sum_{n=0}^\infty \frac{1}{n!}x^n$ and then show that:

• this power series has infinite radius of convergence,
• as such a power series, $E$ is differentiable and equal to its own derivative,
• as a consequence of this, use the MVT to show that $E(a+b)=E(a)E(b)$ for $a,b \in \mathbb{R}$,
• show that $E$ is a strictly increasing bijection from $\mathbb{R}$ to $(0,\infty)$, and deduce the existence of its inverse function $L\,:\, (0,\infty) \to \mathbb{R}$, which has the properties that $L(xy) = L(x)+L(y)$ for $x,y > 0$ and $L'(x) = x^{-1}$.
• define $e = E(1)$, and you now have all the tools needed to follow the original argument.

Your student would need a good foundation in basic analysis (MVT, Intermediate Value Theorem, Inverse Function Theorem, differentiability of power series) to follow this route.

- 6 months, 1 week ago

I just remembered I found another method online to use $\lim_{h\to 0} (1+h)^\frac{1}{h} = e$ to prove $\lim_{h\to 0} \frac{e^h-1}{h}=1,$ but any new answers are welcome.

- 6 months, 1 week ago

When I teach this student, I will be sure to credit you.

- 6 months, 1 week ago

I'm afraid I have some questions that still need to be addressed. If you want to answer them, then that would be great! Define $I(x)=\int_1^x \frac{1}{t}dt.$ I have to prove that $I$ is continuous for $x>0$. This doesn't seem that hard. Then I have to prove $I^{-1}(1)$ exists. I know I have to use the IVT, but I get stuck after that. Any suggestions on how I should show $I^{-1}(1)$ exists?

- 6 months, 1 week ago

I hope you see where this is coming from. In case you don't (which you probably do), let me explain. From your derivation, I get $\lim_{x\to 0+}x^{-1}I(1+x)=1\Rightarrow \lim_{x\to 0+}I((1+x)^\frac{1}{x})=1\Rightarrow I(\lim_{x\to 0+}(1+x)^\frac{1}{x}) = 1.$ So, $\lim_{x\to 0+}(1+x)^\frac{1}{x} = I^{-1}(1).$

- 6 months, 1 week ago

Going this way, clearly $I$ is differentiable with $I'(x) = x^{-1} > 0$ for $x > 0$, so $I$ is strictly increasing. You can then compare areas to show that $I(n) \; \ge \; \sum_{k=2}^n k^{-1} \hspace{2cm} n \ge 2, n \in \mathbb{n}$ and hence $I(x) \to \infty$ as $x\to \infty$. Since $I(xy) = I(x)+I(y)$ (easy variable change) and $I(0)=0$, we have $I(1/x) = -I(x) \to -\infty$ as $x \to \infty$, and hence $I(x) \to -\infty$ as $x \to 0+$. Thus we can use the IVT to deduce that $I\,:\, (0,\infty) \to \mathbb{R}$ is surjective and (since it is strictly increasing) bijective. Thus the Inverse Function Theorem tells us that $I^{-1}=E$ exists, and since $I'(x) \neq 0$ for all $x > 0$, we can deduce that $E'=E$ on the whole of $\mathbb{R}$, and that that $E(x+y) = E(x)E(y)$ for all real $x,y$

- 6 months, 1 week ago

Hmm. well you could also use L'Hospital's Rule once you have $I$ and its properties established to prove $\lim_{h\to 0} (1+h)^\frac{1}{h}=I^{-1}(1)$. I think I would do it that way instead. (Not to knock your proof.) I don't mean to annoy you, but I am quite interested in how to do this without using $I.$ I should've said something before, but I didn't want to overstay my welcome. As far as I can tell, you still haven't outlined a way one can prove the existence of this limt without $I$ or the series definition of $e^x$. However, I think you were saying you could?

- 6 months, 1 week ago

Yes, of course, you can use de l'Hoptal's rule to show that $\lim_{h \to 0+}\tfrac{\ln(1+h)}{h} = \lim_{h \to 0+} \tfrac{1}{1+h} = 1$, and hence $\lim_{h \to 0+} (1+h)^{\frac{1}{h}} = e$. However, this is another proof that uses a number of properties of exponentiation and the logarithm to prove the existence of $e$ - about the same amount of information as was needed for the first proof I gave. In other words, you can provide a hand-waving definition of $e^x$ and the logarithm, and come back and prove the various formulae for $e$. To be honest, that is what most analysis texts do.

That is fine, but it still leaves the fundamental question - what do we mean by $a^b$ anyway? We can define $a^q$ whenever $a>0$ and $q \in \mathbb{Q}$, since $a^{\frac{m}{n}}$ is the $m$th power of the $n$th root of $a$ for any integers $m,n$ with $n > 0$, but the meaning of $a^x$, where $x$ is irrational, is not clear. The only way you can safely define $a^x$ is as $e^{x \ln a}$, and so you have to be able to define the fundamental exponential and logarithm functions before you start talking about general exponentials, or even about the function $2^x$ for all real $x$.

The power series/logarithm approach gives an approach which defines the exponential and the logarithm, either through the power series form of the exponential, or else using the integral definition of the logarithm, and this approach does not assume any prior understanding of exponential or logarithmic functions. From this we can define all exponential and logarithmic properties, and not rely on any underlying heuristics.

- 6 months, 1 week ago

What restricts us from using a series of rational numbers that approach an irrational number to define $a^x$? This is just using limits of sequences to define $a^x,$ right? Although I'm not sure how to prove the limit would exist (converge on one number)... But I bet you could do it. I understand that it would be difficult to conjur a formula for such a sequence of rational numbers that approach a given irrational number or to construct all irrational numbers using these sequences, but how would this be different in principle from taking the limit of a series?

- 6 months, 1 week ago

By wanting $a^x$ to be equal to the limit of $a^{q(n)}$ where $(q(n))$ is a sequence of rationals converging to $x$, you are presupposing that $a^x$ is a continuous function of $x$. Without a definition of the function, you cannot prove continuity. Without continuity, you cannot define the function. Round and round we go...

- 6 months, 1 week ago

Interesting. So you are saying there is no way to prove, using algebra, that $a^{q(n)}$ converges? As you know, I am supposing that $a^x$ for an irrational $x$ be defined as the limit of this sequence. It seems odd that this would not be possible given that we can define transcendental numbers as the limits of sequences.

- 6 months, 1 week ago

I think it is possible to prove that all for all sequences $q(n)$ that approach $x$, $a^{q(n)}$ approaches the same value.

- 6 months, 1 week ago

Sorry. Typo. I meant to type $a^{q(n)}.$ I have fixed it.

- 6 months, 1 week ago

I also think it is possible to prove that after defining $a^x$ for all irrational numbers, that all sequences $a^{s(n)}$ (where $s(n)$ converges to $x$) converge to the same number $a^x$.

- 6 months, 1 week ago

Try replacing "I think" with "here is a proof". Even if you come up with an argument that does not presuppose properties of the exponential and the logarithm, it will be more involved than the approach I suggest. Foundational analysts use the method I have outlined.

- 6 months, 1 week ago

I wonder if a proof like that would make someone famous. Unfortunately, I don't think I can prove it. Better go with the shortcut for now. Anyway, thanks for the discussion and have a nice night/day.

- 6 months, 1 week ago

What you have produced here is quite masterful. I have not seen an equivalent answer anywhere on the internet. This is a subject that needs to be handled with great care. But the power that it produces is worth it. I thank you for participating in this discussion! The series version requires careful analysis techniques that I am not prepared to figure out. I was hoping I would not have to start by defining $\ln(x)$ as $\int_1^x \frac{1}{t}dt.$ But it looks like I will end up going that route. Last I heard, my student was still in pre-calculus, but I haven't talked to her in a while. I'm not sure if she's started calculus yet. I just wanted to be ready. I have further questions if you don't mind. Then to prove that the inverse function of $\int_1^x \frac{1}{t}dt$ is an exponential, just set $x=e^y$ and then use the chain rule to prove $\frac{dy}{dx}=\frac{1}{x}$. Since $\frac{dy}{dx}-\frac{d}{dx}[\int_1^x \frac{1}{t}dt]=\frac{1}{x}-\frac{1}{x}=0,$ we have $y=\int_1^x \frac{1}{t}dt+C.$ Then we can substitute $x=1$ to find $C=0.$ This may sound like a stupid question to the great Mark Hennings, but how do you show $\lim_{h\to 0+}\frac{[(1+h)^\frac{1}{h}]^h-1}{h}=\lim_{h\to 0+} \frac{e^h-1}{h}$? Then I have to prove all exponential (or logarithmic) functions are continuous. Gosh, I have a lot of work to do and notes to take. Forgive me for not knowing these things.

- 6 months, 1 week ago

To be honest, if your student is in pre-calculus, you are probably best not getting too deep at the start - you may confuse her. If you have some nice graphical software, play with the graphs of $a^x$ in the range $2 < x < 3$, particularly if you can show the derivative as well, and show that there is some value between $2$ and $3$ where the derivative switches from being below the function to being above. If you use something like Autograph or Geogebra, you can suggest 4 decimal places of values of $e$ by this method.

First principles differentiation can be used to show that the derivative of $a^x$ w.r.t. $x$ is $K_a \times a^x$ where $K_a$ is a constant that only depends on $a$. Of course, $K_a = \ln a$ is the derivative of $a^x$ at $x=0$. Claim that $e$ is that magic number for which $K_e=1$, and keep going. You can show the sum for $e$ and the limit property at the start as curiosities, and come back to prove them later when she has learned the necessary machinery!

I first learned the fully rigorous approach to defining the exponential and the logarithm at the very end of my first year undergraduate analysis course. At school my teachers did not worry me with issues of existence.

- 6 months, 1 week ago

Has anyone ever told you you are a genius? Thank you so so much. I cannot express my gratitude. I have \$50 to my name right now. I wish there was something I could do to reciprocate. I can use this content for any inquisitive student who asks the right questions.

- 6 months, 1 week ago

Thanks for the appreciation. I am just someone who has been teaching for 40 odd years at both school and university. Glad to share some insights.

- 6 months, 1 week ago