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So lately, when one my friends had asked for help, he showed me this task:

Evaluate the following limit $$\lim_{x\to 0}\frac{x}{2^x-1}$$

The problem is that one is not ought to use L'Hôspital's rule (which yields $\frac{1}{\ln 2}$), because derivatives weren't even introduced yet.

I am sorry that I can't see a way to change this term algebraically and it would be a pleasure if you helped me.

Blue
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Felix Gelbh
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    Well, if we're not going to do this with derivatives (it's rather immediate that the answer has to be $\frac1{\ln 2}$, just from the definition of the derivative applied to $2^x$ at $x = 0$), then we need to know more about the nitty-gritty details. How have you defined the natural logarithm? How have you defined $2^x$? – Arthur Sep 30 '19 at 10:08
  • $\ln(x)=\log_e(x)$ – Felix Gelbh Sep 30 '19 at 10:10
  • That's not a definition. At least not the kind I'm after. If you see something like $\ln(2)$, what does that mean to you? What property does that number have? In other words, I want you to finish the sentence "$\ln 2$ is the unique number such that ..." And "$\ln 2$ is the unique number that is equal to $\log_e(2)$" just isn't going to help me help you figure this out. Also, I would like you to do the same with $2^x$. Mathematical proofs begin and end with definitions, so you need to have those clear in your mind to get anywhere. – Arthur Sep 30 '19 at 10:16
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    Another try: $\ln 2$ is the unique number such that $e^{\ln 2}=2$. – Felix Gelbh Sep 30 '19 at 10:17
  • That's a lot better. I can play the game again with $e^x$, but we have to stop at some point. Now what about $2^x$? – Arthur Sep 30 '19 at 10:18
  • Let $2^x:=\lim_{n\to\infty}2^{q_n}$ if ($q_n$) is a sequence of rational numbers and $\lim_{n\to\infty}q_n=x$. – Felix Gelbh Sep 30 '19 at 10:24
  • The definition of $2^x$ which you mention is the most difficult one. A full development of the theory of exponential and logarithmic functions based on this approach is presented in my post: https://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-3.html?m=0 – Paramanand Singh Sep 30 '19 at 15:59

5 Answers5

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One way

Apply definition of derivative of $2^x$ $$\Longrightarrow \lim_{x\to 0} \frac{2^x-2^0}{x-0}$$ $$=\frac{d}{dx}(2^x)|_0$$ $$=2^0\log 2$$ $$\Longrightarrow \lim_{x\to 0}\frac{x}{2^x-1}=\frac{1}{\log 2}$$

Second way

Apply change of variable, let $2^x=y$ $$\Longrightarrow x=\frac{\log y}{\log 2}$$ Also, as $x\rightarrow 0, y\rightarrow 1$. So, required limit $$=\lim_{y\to 1}\frac{\frac{\log y}{\log 2}}{y-1}$$ $$=\lim_{y\to 1}\frac{\log y}{\log 2 \times (y-1)}$$ Now, apply the well-known fact that $\forall y>0$ $$\frac{y-1}{y}\leq\log y\leq y-1$$ Now, divide both sides by $y-1$ and apply sandwich theorem as $y\rightarrow 1$.

Martund
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  • You should clarify how the last inequality can be derived. – user Sep 30 '19 at 11:53
  • Well, I assumed that OP knows some basic properties of logarithms. Proofs without using derivatives can be found at https://proofwiki.org/wiki/Bounds_of_Natural_Logarithm – Martund Sep 30 '19 at 17:10
  • That's fine but in that case you should specify what kind of derivation you are referring to and possibly give the reference. Thanks for your clarification. – user Sep 30 '19 at 17:17
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Assuming that you know that $(2^x)'=\log(2)\times2^x$, then, in particular, you know that$$\lim_{x\to0}\frac{2^x-1}x=\log(2).$$Therefore,$$\lim_{x\to0}\frac x{2^x-1}=\frac1{\log(2)}.$$

José Carlos Santos
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As $a=e^{\ln a}$ for $a>0$

$$\lim_{x\to0}\dfrac{a^x-1}x=\ln a\cdot\lim_{x\to0}\dfrac{e^{x\ln a}-1}{x\ln a}=?$$

lab bhattacharjee
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Is your friend allowed to use/familiar with derivatives to begin with?

We have $$\displaystyle\lim_{x\to 0}{1\over {2^x - 2^0\over x-0}} = {1\over f’(0)}$$ where $f(x) = 2^x$, so the limit is $\frac 1{ln2}$.

Certainly not a dog
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Since we are not allowed to use derivative we need to start from some well grounded point.

Usually the starting point is the following limit for sequences (wich can be proved by monotonicity theorem):

$$\lim_{n\to \infty}\left(1+\frac1n\right)^n =e$$

which can be easily extended to real functions

$$\lim_{y\to \infty}\left(1+\frac1y\right)^y =e$$

From the latter by $z=\frac1y \to 0$ we easily obtain that

$$\lim_{y\to \infty}\left(1+\frac1y\right)^y =\lim_{z\to 0}\left(1+z\right)^\frac1z=e \implies \lim_{z\to 0}\frac{\log(1+z)}{z}=1$$

and finally by $y=e^x-1 \to 0$ with $x\to 0$

$$\lim_{z\to 0}\frac{\log(1+z)}{z}=1 \implies \lim_{x\to 0}\frac x{e^x-1}=1$$

and then

$$\lim_{x\to 0}\frac{x}{2^x-1}=\lim_{x\to 0}\;\frac1{\log 2}\cdot\frac{x\log 2}{e^{x\log 2}-1}=\frac1{\log 2}\cdot\cdot 1 =\frac1{\log 2}$$

user
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  • OK, thanks for the answer, but it still bugs me how this can be in a German High School text book in Year 11. – Felix Gelbh Sep 30 '19 at 11:58
  • @FelixGelbh Usually the notation $\log 2$ isalso used for $\ln$. I can't explain why. Of course it can be obbained without derivatives but we need to start from some no trivial result. – user Sep 30 '19 at 12:01
  • Excuse me, but can you explain how $\frac{x\log 2}{e^{x\log 2}-1}=\frac{x}{e^x-1}$? @gimusi – Felix Gelbh Sep 30 '19 at 12:21
  • @FelixGelbh Just take as a new variable $y=x\log 2 \to 0$ then $$\frac{x\log 2}{e^{x\log 2}-1}=\frac{y}{e^y-1} \to 1$$ – user Sep 30 '19 at 12:27