It's true that $\lim_{t \to 0}\frac{|x_0 + t|^{p-2}-|x_0|^{p-2}}{t} = 0$?

Question

Let $ f : \mathbb R \longrightarrow \mathbb R $ defined by

$$f(t) = |t|^{p-2}$$

for all $t \in \mathbb R $ with $p >1 $.

I need to show that

$$\lim_{t \to 0}\frac{|x_0 + t|^{p-2}-|x_0|^{p-2}}{t} = 0.$$

My idea is to use the fact that f is convex and therefore worth the inequality

$$f(\frac{a + b}{2}) \leq \frac{f(a) + f(b)}{2}$$

with this we can conclude that for all $\epsilon >0 $ there exists $\delta >0$ such that

$$|x_0 + t|^{p-2}-|x_0|^{p-2} \leq \epsilon |t|$$

for all $ 0 <|t| < \delta$.

How can I proceed to complete this?

Answer 1

Of course this can't be true: otherwise, $f$ would be constant by Lagrange's theorem.

Answer 2

For $x_0>0$, the limit is nothing but the derivative at $x_0$. Since $f'(t)=(p-2)t^{p-1}$, the limit is $(p-2)x_0^{p-1}$. For $x_0<0$ the function is $(-t)^{p-2}$, and a similar reasoning can be made.