1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $(e^h)'=h' c^h=e^h$ and $h'=1$so the limit by Hopital rule is $|lim _{x→0}\frac{e^h}{1}=1$ $\endgroup$ – sirous Jun 26 '18 at 20:23
  • 5
    $\begingroup$ This essentially boils down to clarifying the definition of $e^x$ in your context. $\endgroup$ – Sangchul Lee Jun 26 '18 at 20:49
  • $\begingroup$ Are you saying you have to prove it using the definition you wrote down, or are you looking for other definitions of $e^h$ for which the proof is easier? $\endgroup$ – zhw. Jun 26 '18 at 20:55
  • $\begingroup$ How do you define the exponential? This is straightforward using the $\sum_k {1 \over k!} x^k$ definition. $\endgroup$ – copper.hat Jun 26 '18 at 22:40
  • $\begingroup$ Mr @copper.hat your definition is based on the Taylor series $f(x)= \sum_n {x^n\over n!} \frac{\partial^n f(x)}{\partial x ^n}|_{x=0}$ which also need to know what is the derivative, so I can't use your definition $\endgroup$ – El Mouden Jun 28 '18 at 13:49
8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I especially like solution 1; never saw it before. Although I would say that 2. is easier than referring to 1. $\endgroup$ – zhw. Jun 28 '18 at 18:03
1
$\begingroup$

Logarithm is continuos, 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: $$ \lim_{y\to0}\frac{e^z-1}{z}=1. $$

$\endgroup$
  • 1
    $\begingroup$ But the property of logarithms follows from the derivative of $\ln$ at $1,$ which in turn depends on the thing we're trying to prove $\endgroup$ – zhw. Jun 26 '18 at 20:50
  • $\begingroup$ Wich property? Continuty? $\endgroup$ – zar Jun 27 '18 at 14:08
  • $\begingroup$ "Using a property of logarithms" $\endgroup$ – zhw. Jun 27 '18 at 14:36
  • $\begingroup$ well, that one descends from the definition of logarithm, it has nothing to do with its derivative $\endgroup$ – zar Jun 27 '18 at 15:46
  • $\begingroup$ Which definition of the logarithm you are using? $\endgroup$ – zhw. Jun 27 '18 at 16:31
0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Shouldn't there be more of a rigorous justification of considering each term separately as $n\to \infty$ since the number of terms also goes to $\infty$. $\endgroup$ – Sri-Amirthan Theivendran Jun 26 '18 at 20:25
  • $\begingroup$ @FoobazJohn Here's a more careful justification, if you would like one: $$\begin{align} e^x &=\lim_{n\to\infty} \space\sum_{k=0}^n \frac{n!}{n^k(n-k)!}\frac{x^k}{k!}\\ &=\lim_{m\to\infty }\lim_{n\to\infty} \space\sum_{k=0}^n \frac{m!}{m^k(m-k)!}\frac{x^k}{k!}\\ &=\lim_{n\to\infty} \space\sum_{k=0}^n \lim_{m\to\infty } \frac{m!}{m^k(m-k)!}\frac{x^k}{k!}\\ &=\lim_{n\to\infty} \space\sum_{k=0}^n \frac{x^k}{k!}\\ &=\space\sum_{k=0}^\infty \frac{x^k}{k!}\\ \end{align}$$ $\endgroup$ – Frpzzd Jun 26 '18 at 20:42
  • 4
    $\begingroup$ You say something approaches $1$ by just repeating the claim in different notation? More importantly, you are using the dominated convergence theorem without saying so. $\endgroup$ – zhw. Jun 26 '18 at 20:42
  • 1
    $\begingroup$ The first double limit in your comment doesn't make sense to me. $k>m$ is possible. what is $(m-k)!$ when $k>m?$ $\endgroup$ – zhw. Jun 26 '18 at 20:47
  • $\begingroup$ Like @zhw. mentions your approach is a classic case of monotone convergence theorem and the proof of the theorem is not difficult but not trivial too. Another simpler approach is to show that $$\left(1+\frac{x}{n}\right)^n\leq\sum_{k=0}^{n}\frac{x^k}{k!}\leq \left(1-\frac{x}{n}\right)^{-n},0<x<n$$ via binomial theorem and then take limits as $n\to\infty$. $\endgroup$ – Paramanand Singh Jun 27 '18 at 6:27
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ FYI, the assertion that $$\lim_{h\to0} e^h=\lim_{h\to0}\left(1+h\right)$$ is true and quite unrelated to the fact that $$\lim_{h\to 0} \, \frac{e^h-1}{h}=1$$ Note for example that one also has $$\lim_{h\to0} e^h=\lim_{h\to0}\left(1+7h\right)$$ and that you would probably not deduce from this other true fact that $$\lim_{h\to 0} \, \frac{e^h-1}{h}=7$$ $\endgroup$ – Did Aug 8 '18 at 4:46
-1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ I think he try to prove that $De^x=e^x$(I didn't downvote) $\endgroup$ – Holo Jun 26 '18 at 20:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.