Prove that $\left|\frac{x^p-1}{p}\right|\leq x+|\ln(x)|$ for all $x\in(0,\infty)$ and for all $p\in(0,1)$ So far I have shown that $$\displaystyle \lim\limits_{p\to 0^+}\frac{x^p-1}{p}=\lim\limits_{p\to 0^+}\frac{e^{p\ln(x)}-1}{p}=\ln(x)\lim\limits_{p\to 0^+}\frac{e^{p\ln(x)}-1}{p\ln(x)} =$$
(L'Hopital)
$$=\ln(x) \lim\limits_{p\to 0^+}\frac{\ln(x)e^{p\ln(x)}}{\ln(x)}=\ln(x).$$
But I don't know if it helps.
A hint would be more appreciated then the whole answer.

If $x \geq 1$, then $LHS = \dfrac{-1+x^p}{p}, RHS = x + \ln x$. Thus we prove:
$x^p-1 \leq px +p\ln x\iff f(x)=-1+x^p-px -p\ln x \leq 0$. We have: $f'(x) = px^{p-1}-p-\dfrac{p}{x}=\dfrac{p}{x}\cdot (x^p-x-1)<0\to f(x) \leq f(1)=-p<0$. I leave the other part for you as a practice.
