How to prove that $\lim_{h\to 0} \, \frac{e^h-1}{h}$

Question

when I try to show that the derivative of exp function is $$\frac{\partial e^t}{\partial t}=e^t$$ I need to prove firstly that this limit is equal to 1.

$$\lim_{h\to 0} \, \frac{e^h-1}{h}=1$$

If you try to use L'Hospital's rule then you need firstly to know what is the derivative of exp function that we want to prove it.

If you want to use approximation of $e^\epsilon \simeq1+\epsilon$ you well found the answer '$1$' but I don't want to use this approximation because it based on the Taylor series that required also to know the derivative of exp function that we don't have it yet

The only way that I can prove it, is with the definition of $e$ number $$ \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e $$ so I ask if there are another way to show this limit.

Answer 1

Thank you all for your answers.

But my question is about if there are any other method without use any of the definition or derivative.

As I said in my Question I already know the "only way" to calculate this limit, $$\lim_{h\to 0} \, \frac{e^h-1}{h}=1$$ "the only way" i mean : using this definition

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

by substituting $x=1/h$ you can rewrite it as

$$e=\lim_{h\to0}\left(1+h\right)^{1/h}$$

then : $$\lim_{h\to0} e^h=\lim_{h\to0}(\left(1+h\right)^{1/h})^h =\lim_{h\to0}\left(1+h\right) $$ so you can easily prove it the limit equal to one

$$\lim_{h\to 0} \, \frac{e^h-1}{h}=1$$

so we can say that : this limit is just redefine of the definition of the $e$ number

Answer 2

We have that for $x\to \infty$

$$ \left(1+\frac{1}{x}\right)^x\le e \le \left(1+\frac{1}{x}\right)^{x+1}$$

thus for $h\to 0^+$

$$ \left(1+h\right)^\frac1h \le e \le \left(1+h\right)^{\frac1h +1}$$

and thus

$$ 1 \le \frac{e^h-1}h \le \frac{\left(1+h\right)^{1+h}-1}{h} =\frac{\left(1+h\right)\left(1+h\right)^{h}-1}{h}\le \frac{\left(1+h\right)\left(1+h^2\right)-1}{h}=\frac{h+h^2+h^3}{h}\to 1$$

Answer 3

Perhaps not the most efficient method, but rather crafty (in my opinion).

You can use a Taylor Series for $e^x$ without knowing its derivatives. The Taylor Series of $e^x$ can be proven using only the definition of $e$ and the Binomial Theorem. Notice that $$e^x=\lim_{n\to\infty} \bigg(1+\frac{1}{n}\bigg)^{nx}$$ and $$e^x=\lim_{n\to\infty} \bigg(1+\frac{x}{n}\bigg)^{n}$$ Then, using the Binomial Theorem, we have $$\begin{align} e^x &=\lim_{n\to\infty} \space\sum_{k=0}^n \binom{n}{k}\bigg(\frac{x}{n}\bigg)^k\\ &=\lim_{n\to\infty} \space\sum_{k=0}^n \frac{n!}{k!(n-k)!}\bigg(\frac{x}{n}\bigg)^k\\ &=\lim_{n\to\infty} \space\sum_{k=0}^n \frac{n!}{n^k(n-k)!}\frac{x^k}{k!}\\ \end{align}$$ As $n\to\infty$, each coefficient $$\frac{n!}{n^k(n-k)!}$$ approaches $1$, since $$\lim_{n\to\infty}\frac{n!}{n^k(n-k)!}=1$$ for all nonnegative integers $k$. Thus, we end up with the desired result of $$e^x=\lim_{n\to\infty} \space\sum_{k=0}^n \frac{x^k}{k!}=\sum_{k=0}^\infty \frac{x^k}{k!}$$ Then you can use this "Taylor Series" just like you wanted to.

Answer 4

Logarithm is continuous, so from $$ \lim_{x\to\pm\infty}\left(1+\frac{1}{x}\right)^x=e $$ you can obtain $$ \lim_{x\to\pm\infty}\ln\left(1+\frac{1}{x}\right)^x=\ln e=1. $$ Using a property of logarithms: $$ \lim_{x\to\pm\infty}x\ln\left(1+\frac{1}{x}\right)=1, $$ and, by substituting $y=1/x$, you have $$ \lim_{y\to0}\frac{\ln(1 + y)}{y}=1. $$ Now, substituting $z=\ln(1 + y)$, so that $y=e^z - 1$, you have $$ \lim_{y\to0}\frac{z}{e^z - 1}=1. $$ Now, taking reciprocals: $$ \lim_{y\to0}\frac{e^z - 1}{z}=1. $$

Answer 5

As I pointed out in my comment, the answer depends on which definition you are using.

Solution 1. Here we define $e^x = \lim_{n\to\infty}\left(1 + \frac{x}{n}\right)^n$. (We assume the existence of this limit is already established.) Using the binomial theorem, for $n \geq 2$ and $|x| < 1$ we obtain the following simple estimate:

$$ \left| \left(1 + \frac{x}{n}\right)^n - 1 - x \right| = \left| \sum_{k=2}^{n} \binom{n}{k} \frac{x^k}{n^k} \right| \leq \sum_{k=2}^{n} |x|^k \leq \frac{|x|^2}{1-|x|}. $$

This bound remains true as we let $n\to\infty$, so it follows that

$$ \left| \frac{e^x - 1}{x} - 1\right| = \frac{\left|e^x - 1 - x\right|}{|x|} \leq \frac{|x|}{1-|x|} $$

and by the squeezing lemma as $x\to0$ the desired conclusion follows.

Solution 2. In case $e^x$ is defined as $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$, the above argument carries over in almost identicaly way to yield a proof.

Solution 3. In case $e^x$ is defined as the inverse function of $x \mapsto \int_{1}^{x}\frac{dt}{t}$, it follows from the inverse function theorem.

Solution 4. Perhaps the most pesky case is where the exponentiation is defined by the process of extending rational exponents to reals and $e$ is simply defined as $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$. In this case, one can introduce two functions, $\log(x) := \int_{1}^{x}\frac{dt}{t}$ and its inverse, and observe that

• $\log(xy) = \int_{1}^{x} \frac{dt}{t} + \int_{x}^{xy} \frac{dt}{t} = \log(x) + \log(y)$ and hence $\exp(x+y) = \exp(x)\exp(y)$.

• $\log\left(\left(1+\frac{1}{n}\right)^n\right) = n\log\left(1+\frac{1}{n}\right) \to 1$ as $n\to\infty$ and $\log$ is continuous, so $\log(e) = 1$. In other words, $\exp(1) = e$.

• Using this, it is routine to check that $\exp(k) = e^k$ for all integers $k$ and then $\exp(r) = e^r$ for all rationals $r$. Hence for any real $x$,

  $$ e^x = \inf\{ e^r : r > x \text{ and } r \in \mathbb{Q} \} = \inf\{ \exp(r) : r > x \text{ and } r \in \mathbb{Q} \} = \exp(x) $$

  by the continuity of $\exp$.

Now the conclusion follows by Solution 3.

Answer 6

Since $(1+\frac{x}{n})^{n+1}$ is decreasing for $x\leq 1$ you have $$1\leq \frac{e^x-1}{x}\leq \frac{1}{x}[(1+\frac{x}{n})^{n+1}-1]$$