Why doesn't $\frac{d(e^x)}{dx}=\frac{0}{0}$?

Question

$$\frac{d(e^x)}{dx}=\lim_{h \mapsto 0}\frac{e^{x+h}-e^x}{h}$$

but

$$\begin{align}h \mapsto 0 \implies \frac{e^{x+h}-e^x}{h}&\mapsto \frac{e^{x+0}-e^x}{0} \\ &=\frac{0}{0}\end{align}$$

What's gone wrong?

Answer 1

The derivative of $f(x)$ is defined as $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.$$You can rewrite a limit of a ratio as a ratio of limits, if the denominator limit is nonzero, but of course in this case it isn't. When we write $$L=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$we mean only that $\frac{f(x+h)-f(x)}{h}$ is as close as you want to $L$, for sufficiently small $h\ne0$. The behaviour "at" $0$ isn't covered in this definition of a limit.

Answer 2

What you write is not so stupid. It is correct that both enumerator and denominator tend to zero. Let us consider a general expression like $$ \lim_{x\to 0}\frac{u(x)}{v(x)}, $$ where $u$ and $v$ are continuous functions with $u(0) = 0$ and $v(0) = 0$. When you don't have more information on $u$ and $v$, the expression is worthless because the limit might not even exist. It could be anything. It strongly depends on the behaviour of $u$ and $v$. Let me give you a few simple examples:

(1) Let $u(x) = x$ and $v(x) = x^2$. Then actually $\frac{u(x)}{v(x)} = \frac 1x$ and the limit does not exist (it is actually $\infty$, but we don't consider this as a limit).

(2) $u(x) = 2x$, $v(x) = x$. Again, both tend to zero, but $\frac{u(x)}{v(x)} = 2$ and so the limit exists and is clearly $2$.

(3) $u(x) = x^2$, $v(x) = x$. Then $\frac{u(x)}{v(x)} = x$ and the limit exists and is zero.

In the case of the derivative of the exponential function, the existence of the limit is not that obvious as in the above examples. You have to elaborate a little more there.

Answer 3

Well, I'd take the definition of $e^x = 1 + x+ x^2/2! + x^3/3! + \ldots$ as Taylor series expansion.

Then the quotient $\frac{1}{h}(e^{x+h}-e^x)$ becomes

$ \frac{1}{h}[(1 + (x+h)+ (x+h)^2/2! + (x+h)^3/3! + \ldots) - ( 1 + x+ x^2/2! + x^3/3! + \ldots)]$

$= \frac{1}{h}[h + (2xh+h^2)/2! + (3x^2h+3xh^2+h^3)/3! + \ldots]$

Multiplying with $\frac{1}{h}$ and using $h\rightarrow 0$ gives

$= 1 + x + x^2/2! + x^3/3! + \ldots =e^x.$

Note that in such proofs it is always important to divide by $h$ (i.e., get rid of the denominator!) and then form the limit $h\rightarrow 0$ (i.e., $h$ becomes small and ugly as my teacher sometimes pointed out).