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Can anybody show me how $\lim \limits_{h \to 0} \dfrac{e^{h} - 1}{h} = 1$?

Daniel W. Farlow
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Philippe Sussholz
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    what is your definition of $e?$ – abel Apr 06 '15 at 16:04
  • What abel said. To do this we need your definition of $e^h$. – GEdgar Apr 06 '15 at 16:05
  • This is actually a reasonable way to define $e$: the unique real number with this property. If you use $e=\lim_{n \to \infty} (1+1/n)^n$, then you have some work with binomial expansion to check this result. Notably this method is somewhat hard because it requires you to carefully define what $x^y$ means when $y$ is irrational. – Ian Apr 06 '15 at 16:08
  • There are many definitions of $e^x$: inverse of natural log, where $\log x = \int_1^x dt/t$, or solution of $y'=y$ with $y(0)=1$, or Taylor series, or limit of $(1+x/n)^n$, ... Depending on yours, the answer to your question may be a bit hard, or almost trivial, or itself a definition. – Jean-Claude Arbaut Apr 06 '15 at 16:27
  • Why not just use l'hopitals rule? – Matthew Cassell Apr 06 '15 at 16:29
  • @Mattos, that would be circular. To use l'Hopital, you need to know the derivative of $e^h.$ But the question asked is how to compute the derivative of $e^h$ at $0.$ – zhw. Apr 06 '15 at 17:26

5 Answers5

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If your definition of $$e^{x} = \sum_{n=0}^{\infty}{\frac{x^{n}}{n!}}$$ Then: $$\frac{e^{h}-1}{h} = \frac{1}{h}\left( \sum_{n=0}^{\infty}{\frac{h^{n}}{n!}}-1 \right)=\frac{1}{h}\left( \sum_{n=2}^{\infty}{\frac{h^{n}}{n!}}+1+h-1 \right) = \frac{1}{h}\left( \sum_{n=2}^{\infty}{\frac{h^{n}}{n!}}+h \right)=\frac{h}{h}\left( \sum_{n=2}^{\infty}{\frac{h^{n-1}}{n!}}+1 \right)= h^{2} \sum_{n=2}^{\infty}{\frac{h^{n-2}}{n!}}+1$$ Notice: $$\sum_{n=2}^{\infty}{\frac{h^{n-2}}{n!}}=\sum_{n=0}^{\infty}{\frac{h^{n}}{(n+2)!}}\leq \sum_{n=0}^{\infty}{\frac{h^{n}}{n!}}=e^h < \infty$$ So: $$\mathop{lim}_{h \rightarrow 0} \frac{e^{h}-1}{h} = \mathop{lim}_{h \rightarrow 0} (h^{2}e^h+1)=0+1=1$$

abcdef
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You can also use the fact that $\frac{e^{h}-1}{h}=\frac{e^{h}-e^{0}}{h-0}$, and as h goes to zero, it gives you the value of the derivative of $e^{x}$ at 0 (since the exponential function is differentiable for all x) and since its equal to its derivative, the limit is $e^{0}=1$

mich95
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I'm guessing you want to do this to prove that $(e^x)'=e^x$. So $e^h=\sum_{n=0}^{\infty} \dfrac{h^n}{n!}$. Then $e^h-1=\sum_{n=1}^{\infty}\dfrac{h^n}{n!}$.

Thus $\dfrac{e^h-1}{h}=\sum_{n=1}^{\infty} \dfrac{h^{n-1}}{n!}=1+\sum_{n=2}^{\infty} \dfrac{h^{n-1}}{n!}$. Taking the limit as $h\to 0$, this last term goes to $0$.

Moya
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Note that $$ \lim_{h\to0}\frac{e^{h}-1}{h}=\lim_{h\to0}\frac{e^{h}-e(0)}{h-0} $$

which is, by definition, the derivative of $e^{x}$evaluated at $0$. Since $(e^{x})'=e^{x}$ the limit is $e^{0}=1$.

Belgi
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I found something in a high school book that was simple and convincable. (No nned for l'Hopital rule , or the Taylor series) e = lim h→0 (1 + h ) ^ 1/h lim h→0 e^h = lim h→0 (1+h) ^ h/h = lim h→0 1+h (I guess this makes sense) lim h→0 (e^h -1 ) / h = 1

Philippe Sussholz
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