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, ee, as limh0(1+h)1h.\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
3 months, 1 week ago

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Start by proving that x12x2<ln(1+x)<xx - \tfrac12x^2 < \ln(1+x) < x for all x>0x > 0.

  • If f(x)=xln(1+x)f(x) = x - \ln(1+x), then f(x)=x1+x>0f'(x) = \tfrac{x}{1+x} > 0 for all x>0x > 0 and f(0)=0f(0) =0. The MVT gives f(x)>0f(x) >0 for x>0x > 0.
  • If g(x)=ln(1+x)x+12x2g(x) = \ln(1+x)-x+\tfrac12x^2, then g(x)=x21+x>0g'(x) = \tfrac{x^2}{1+x} > 0 for all x>0x > 0, and g(0)=0g(0)= 0. The MVT gives g(x)>0g(x) > 0 for x>0x > 0.

Then 12x<x1ln(1+x)1<0-\tfrac12x < x^{-1}\ln(1+x) - 1 < 0 for x>0x > 0, and hence limx0+x1ln(1+x)=1\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 ee, unless you take the approach below...

Define exe^x by its power series E(x)  =  n=01n!xn 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, EE 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)E(a+b)=E(a)E(b) for a,bRa,b \in \mathbb{R},
  • show that EE is a strictly increasing bijection from R\mathbb{R} to (0,)(0,\infty), and deduce the existence of its inverse function L:(0,)RL\,:\, (0,\infty) \to \mathbb{R}, which has the properties that L(xy)=L(x)+L(y)L(xy) = L(x)+L(y) for x,y>0x,y > 0 and L(x)=x1L'(x) = x^{-1}.
  • define e=E(1)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.

Mark Hennings - 1 month, 1 week ago

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I just remembered I found another method online to use limh0(1+h)1h=e\lim_{h\to 0} (1+h)^\frac{1}{h} = e to prove limh0eh1h=1,\lim_{h\to 0} \frac{e^h-1}{h}=1, but any new answers are welcome.

James Wilson - 1 month, 1 week ago

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

James Wilson - 1 month, 1 week ago

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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)=1x1tdt.I(x)=\int_1^x \frac{1}{t}dt. I have to prove that II is continuous for x>0x>0. This doesn't seem that hard. Then I have to prove I1(1)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 I1(1)I^{-1}(1) exists?

James Wilson - 1 month, 1 week ago

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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 limx0+x1I(1+x)=1limx0+I((1+x)1x)=1I(limx0+(1+x)1x)=1.\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, limx0+(1+x)1x=I1(1).\lim_{x\to 0+}(1+x)^\frac{1}{x} = I^{-1}(1).

James Wilson - 1 month, 1 week ago

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Going this way, clearly II is differentiable with I(x)=x1>0I'(x) = x^{-1} > 0 for x>0x > 0, so II is strictly increasing. You can then compare areas to show that I(n)    k=2nk1n2,nn I(n) \; \ge \; \sum_{k=2}^n k^{-1} \hspace{2cm} n \ge 2, n \in \mathbb{n} and hence I(x)I(x) \to \infty as xx\to \infty. Since I(xy)=I(x)+I(y)I(xy) = I(x)+I(y) (easy variable change) and I(0)=0I(0)=0, we have I(1/x)=I(x)I(1/x) = -I(x) \to -\infty as xx \to \infty, and hence I(x)I(x) \to -\infty as x0+x \to 0+. Thus we can use the IVT to deduce that I:(0,)RI\,:\, (0,\infty) \to \mathbb{R} is surjective and (since it is strictly increasing) bijective. Thus the Inverse Function Theorem tells us that I1=EI^{-1}=E exists, and since I(x)0I'(x) \neq 0 for all x>0x > 0, we can deduce that E=EE'=E on the whole of R\mathbb{R}, and that that E(x+y)=E(x)E(y)E(x+y) = E(x)E(y) for all real x,yx,y

You can start with the logarithm, or you can start with the exponential. It does not matter which.

Mark Hennings - 1 month, 1 week ago

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Hmm. well you could also use L'Hospital's Rule once you have II and its properties established to prove limh0(1+h)1h=I1(1)\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. 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 II or the series definition of exe^x. However, I think you were saying you could?

James Wilson - 1 month, 1 week ago

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Yes, of course, you can use de l'Hoptal's rule to show that limh0+ln(1+h)h=limh0+11+h=1\lim_{h \to 0+}\tfrac{\ln(1+h)}{h} = \lim_{h \to 0+} \tfrac{1}{1+h} = 1, and hence limh0+(1+h)1h=e\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 ee - 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 exe^x and the logarithm, and come back and prove the various formulae for ee. 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 aba^b anyway? We can define aqa^q whenever a>0a>0 and qQq \in \mathbb{Q}, since amna^{\frac{m}{n}} is the mmth power of the nnth root of aa for any integers m,nm,n with n>0n > 0, but the meaning of axa^x, where xx is irrational, is not clear. The only way you can safely define axa^x is as exlnae^{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 2x2^x for all real xx.

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.

Mark Hennings - 1 month, 1 week ago

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@Mark Hennings What restricts us from using a series of rational numbers that approach an irrational number to define axa^x? This is just using limits of sequences to define ax,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?

James Wilson - 1 month, 1 week ago

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

Mark Hennings - 1 month, 1 week ago

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@Mark Hennings Interesting. So you are saying there is no way to prove, using algebra, that aq(n)a^{q(n)} converges? As you know, I am supposing that axa^x for an irrational xx 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.

James Wilson - 1 month, 1 week ago

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@Mark Hennings I think it is possible to prove that all for all sequences q(n)q(n) that approach xx, aq(n)a^{q(n)} approaches the same value.

James Wilson - 1 month, 1 week ago

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@James Wilson Sorry. Typo. I meant to type aq(n).a^{q(n)}. I have fixed it.

James Wilson - 1 month, 1 week ago

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

James Wilson - 1 month, 1 week ago

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@James Wilson 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.

Mark Hennings - 1 month, 1 week ago

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@Mark Hennings 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.

James Wilson - 1 month, 1 week ago

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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)\ln(x) as 1x1tdt.\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 1x1tdt\int_1^x \frac{1}{t}dt is an exponential, just set x=eyx=e^y and then use the chain rule to prove dydx=1x\frac{dy}{dx}=\frac{1}{x}. Since dydxddx[1x1tdt]=1x1x=0,\frac{dy}{dx}-\frac{d}{dx}[\int_1^x \frac{1}{t}dt]=\frac{1}{x}-\frac{1}{x}=0, we have y=1x1tdt+C.y=\int_1^x \frac{1}{t}dt+C. Then we can substitute x=1x=1 to find C=0.C=0. This may sound like a stupid question to the great Mark Hennings, but how do you show limh0+[(1+h)1h]h1h=limh0+eh1h\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.

James Wilson - 1 month, 1 week ago

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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 axa^x in the range 2<x<32 < x < 3, particularly if you can show the derivative as well, and show that there is some value between 22 and 33 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 ee by this method.

First principles differentiation can be used to show that the derivative of axa^x w.r.t. xx is Ka×axK_a \times a^x where KaK_a is a constant that only depends on aa. Of course, Ka=lnaK_a = \ln a is the derivative of axa^x at x=0x=0. Claim that ee is that magic number for which Ke=1K_e=1, and keep going. You can show the sum for ee 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.

Mark Hennings - 1 month, 1 week ago

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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.

James Wilson - 1 month, 1 week ago

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@James Wilson 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.

Mark Hennings - 1 month, 1 week ago

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