Consider the function $f:x \neq 0 \mapsto \frac x {\log |x|}, 0 \mapsto 0$. What is its derivative at $x = 0$?
I was only able to solve this using L'Hopital's rule. I request:
- Verification or critique of my solution
- Is it possible to solve without L'Hopital's rule?
Proof:
$$\begin{align} \lim_{h \to 0^+} \frac {\frac h {\log h} - 0}{h} &= \lim_{h \to 0^+} \frac 1 {\log h} \\ &= \lim_{h \to 0^+} \frac 0 {1/h} \tag{by L'Hopital's Rule} \\ &= 0. \end{align}$$
Similarly, $$\begin{align} \lim_{h \to 0^-} \frac {\frac h {\log -h} - 0}{h} &= \lim_{h \to 0^-} \frac 1 {\log -h} \\ &= \lim_{h \to 0^-} \frac 0 {1/h} \tag{by L'Hopital's Rule} \\ &= 0. \end{align}$$
Since the left and right limits are equal, the limit is defined, and is $0$.
Update
With help from Thomas Andrews' comment, I replace my (mis?)application of L'Hopital's Rule with the following lemma:
Lemma: Let $g: \mathbb R \to \mathbb R$ be a function such that for all $R$ there exists $\delta > 0$ such that for all $h \neq 0 \in [-\delta, \delta], g(k+h) \geq R$. Then $\lim_{x \to k} \frac 1 {g(x)} = 0.$
Proof: For any $\epsilon > 0 $, choose $R > 1/\epsilon$. Then for all $h \neq 0 \in [-\delta, \delta],$ $$\frac 1 {|g(k+h)|} \leq \frac 1 R < \epsilon.$$