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Given $ƒ(x) = e^x$, verify that $$\lim_{h\to 0}\frac{e^{x+h} – e^x}{h} = e^x$$ and explain how this illustrates that $f'(x) = \ln e \cdot f(x) = f(x)$.

Em.
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Michelle
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  • Formatting tips here. – Em. Apr 29 '16 at 03:38
  • It's a little complicated to do this if we don't know exactly the definition of the exponential you're using, if any. As for the "explain how this illustrates...", the limit is simply the definition of the derivative of $f$ at $x$. – Luiz Cordeiro Apr 29 '16 at 03:51
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    @RRL They are a totally different questions. Even the equation is different. – Kate.K Apr 29 '16 at 04:11

3 Answers3

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A proper answer to this question is fully dependent on the proper definition of symbol $e^{x}$. And further a proper definition of $e^{x}$ can not be given by just defining $e$.

One definition of $e^{x}$ is given by $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{1}$$ Although it is normally self-evident because of the symbol $n$, I want to emphasize that the $n$ in the above definition is a positive integer. Based on this definition it can be proven that $$e^{x + y} = e^{x}\cdot e^{y}\tag{2}$$ and $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1\tag{3}$$ Both the equations $(2)$ and $(3)$ are proved in this answer based on definition $(1)$. Using these equations $(2), (3)$ it is easy to prove that $$\frac{d}{dx}\,e^{x} = e^{x}\tag{4}$$ Second part of the question requires us to define the symbol $\ln$ (or $\log$ properly). One such definition is that if $e^{y} = x$ then $y = \ln x$ and then $\ln e = 1$ and this solves the second part of your question.

Community
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Paramanand Singh
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Recall that all we need to show is that $\lim_{h\to 0} \frac{e^h - 1}{h} = 1$. The only things we need in this case are the definition of $e$ given by $e^h = \lim_{n\to \infty}\left ( 1 + \frac{h}{n} \right )^n$, and the binomial theorem. Observe that

\begin{eqnarray*} \lim_{h\to 0} \frac{e^h -1}{h} & = & \lim_{h\to 0} \lim_{n\to \infty} \frac{1}{h} \left ( 1 + \frac{h}{n} \right )^n - \frac{1}{h} \\ & = & \lim_{h\to 0}\lim_{n\to \infty} \frac{1}{h} \sum_{k=0}^n \binom{n}{k}\left ( \frac{h}{n}\right )^k - \frac{1}{h}. \end{eqnarray*}

If we factor out $\frac{1}{h}$ from both terms, we see that we can simply start the summation from the $k=1$ term, hence we have

\begin{eqnarray*} \lim_{h\to 0} \frac{e^h-1}{h} & = & \lim_{h\to 0} \lim_{n\to \infty} \frac{1}{h} \sum_{k=1}^n \binom{n}{k} \left ( \frac{h}{n} \right )^k \\ & = & \lim_{h\to0} \lim_{n\to \infty}\frac{1}{h} \left (n \frac{h}{n} + \frac{n(n-1)}{2}\frac{h^2}{n^2} + \frac{n(n-1)(n-2)}{3!}\frac{h^3}{n^3} + \ldots + \frac{h^n}{n^n} \right ) \\ & = & \lim_{h\to 0} \lim_{n\to\infty} \left( 1 + \binom{n}{2}\frac{h}{n} + \binom{n}{3} \frac{h^2}{n^3} + \ldots + \frac{h^{n-1}}{n^n} \right ). \end{eqnarray*}

No matter what the form of the series is in the limit we'll have that there are terms in various powers of $h$, which in the limit as $h\to 0$ will all vanish. The only term that remains is the value of $1$ at the beginning of the summation.

Mnifldz
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    While your answer is intuitively correct, it does lack rigor. First of all from the definition $e^{h} = \lim_{n \to \infty}(1 + (h/n))^{n}$ it is not obvious that $e^{x + y} = e^{x}\cdot e^{y}$. Next in the final part of your answer evaluating a double limit of type $\lim_{h \to 0}\lim_{n \to \infty}$ is not easy and needs more work to show that the limit is $1$. See my answer http://math.stackexchange.com/a/541330/72031 for these points. – Paramanand Singh Apr 29 '16 at 06:53
  • It is far to be clear that the double limit is $1$ because almost all terms are undeterminated forms. – Masacroso Apr 29 '16 at 06:57
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Using rules of exponentials,

$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = \lim_{h \rightarrow 0} \frac{e^xe^h-e^x}{h}$$ $$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = e^x \lim_{h \rightarrow 0} \frac{e^h-1}{h}$$

This has a limit of $1$ (although I'm not sure how to show this without using L'Hospital's rule. Therefore,

$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = e^x$$

user1543042
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    L'Hopital's rule automatically assumes that you know how to take the derivative of $e^x$, but this is what the OP is trying to prove using first principles. – Mnifldz Apr 29 '16 at 06:07