Show that $\lim_{p \to 0} \mathbb {E} [(X^p-1)/p ] = \mathbb{E} [\log X]$ 
Show that $\lim_{p \to 0} \mathbb {E} [(X^p-1)/p ] = \mathbb{E} [\log X]$

I am trying to show this equality by using the Dominated Convergence Theorem. Here $X$ is a non-negative random variable and $\mathbb {E}[X^p] < \infty$ for some $p >0$. I already found that $\lim_{p\to 0}\frac{X^p-1}{p} = \log X$. However, I can't find an uniform bound of $\frac{X^p-1}{p}$ for all $p$. Can anyone give me some hint on how to manipulate the sequence $\left\{\frac{X^p-1}{p}\right\}_{p>0}$
 A: I first claim that if $0<p<q<1$, then $\frac{t^p-1}{p}\leq \frac{t^q-1}{q}$ for all $t\geq 0$. To prove this claim, define $\phi(t)=\frac{t^q-1}{q}-\frac{t^p-1}{p}$. Then
$$ \phi^{\prime}(t)=t^{q-1}-t^{p-1}=t^{p-1}(t^{q-p}-1)$$
which has its unique root at $t=1$. Since $\phi^{\prime\prime}(1)=q-p>0$, this is a minimum, and since $\phi(1)=0$ it follows that $\phi(t)\geq 0$ for all $t\geq 0$.
Now that we have the claim, fix some $0<q<1$ such that $\mathbb{E}[X^q]<\infty$. Our claim shows that the family $\{\frac{X^p-1}{p}\}$ is monotone decreasing in $p$ as $p\to 0$, and since $\frac{X^p-1}{p}\leq \frac{X^q-1}{q}$ for all  $0<p<q$ we can let $p\to 0$ along any sequence and use the decreasing variant of the monotone convergence theorem to conclude that
$$ \lim_{p\to 0}\mathbb{E}\Big[\frac{X^p-1}{p}\Big]=\mathbb{E}[\log X]$$
A: Hint: 


*

*Show that the mapping $(0,\infty) \ni p \mapsto f(p) := \frac{x^p-1}{p}$ is non-decreasing for any $x>0$. (Hint: Use the elementary estimate $$\log(y) +1 \leq y, \qquad y>0$$ to deduce that the derivative $f'$ is non-negative.)

*Conclude that $$\log x = \inf_{p>0} \frac{x^p-1}{p}$$ for all $x>0$.

*Apply the monotone convergence theorem.

