Proof that a continuous function is differentiable

Question

I have been introduced to the function that is 0 for $t\leq 0$ and $e^{-1/t}$ for positive $t$. My lecturor's proof that this function is differentiable (even infinitely, but irrelevant here) involves it being continuous everywhere and differentiable everywhere except maybe at 0. Now, its differential tends to the same limit on both sides of 0, and from that he concludes. It seems quite believable to me that it is differentiable at 0 but I can't seem to feel totally convinced. Is this easy to prove ? Thanks

Answer 1

Differentiability on the left is trivial:for $h>0$, $$ \frac{f(0-h)-f(0)}{h} = \frac{0-0}{h} = 0 \xrightarrow[h\to 0^+]{} 0 $$

Differentiability on the right $h>0$, $$ \frac{f(0+h)-f(0)}{h} = \frac{e^{-1/h}-0}{h} = \frac{1}{h}e^{-\frac{1}{h}} $$ To compute this limit as $h\to 0^+$, observe that by doing the substitution $x\stackrel{\rm def}{=} \frac{1}{h}$ this is equivalent to computing the limit of $x e^{-x}$ when $x\to+\infty$. There are numerous proofs of this on this website (and it can be considered a standard fact) that "exponential beats powers:" $\lim_{x\to+\infty}x e^{-x} = 0$. Therefore, $$ \frac{f(0+h)-f(0)}{h} \xrightarrow[h\to 0^+]{} 0. $$

Both one-sided derivatives exist at $0$, and they are equal (to $0$): thus, $f$ is differentiable at $0$, and $f'(0)=0$.