Evaluating the limit: $\lim_{x\rightarrow -1^+}\sqrt[3]{x+1}\ln(x+1)$ I need to solve this question: $$\lim_{x\rightarrow -1^+}\sqrt[3]{x+1}\ln(x+1)$$
I tried the graphical method and observed that the graph was approaching $0$ as $x$ approached $-1$ but I need to know if there's a way to calculate this.

Standard fact: 
For any real $\alpha,\beta>0$,
$$\lim_{x\to 0^+}x^\alpha \lvert\ln x\rvert^\beta=0.$$
Now just set $u=x+1$ and appropriate $\alpha$  and $\beta$.
Proof of the standard fact:
We prove first that for any $\alpha,\, \beta>0$, $\,\ln^\beta x=_\infty o(x^\alpha)$.
It is easy to prove that, for all $x>0$, $\ln x <2\sqrt x$. We deduce that for any $r>0$, $\;\ln(x^r)<2x^{r/2}$, whence $\ln x<\dfrac2r\, x^{r/2}$, so that
$$\frac{\ln x}{x^r}<\frac 2{r x^{r/2}}\enspace\text{whence}\enspace\frac{\ln^\beta x}{x^{r\beta}}<\frac{2^\beta}{r^\beta x^{r\beta/2}} $$
If we choose $r=\alpha/\beta$, we obtain:
$$\frac{\ln^\beta x}{x^\alpha}<\Bigl(\frac{2\beta}{\alpha}\Bigr)^\beta \frac 1{x^{\alpha/2}}, \enspace\text{thus}\quad \lim_{x\to\infty}\frac{\ln^\beta x}{x^\alpha}=0$$
The limit at $\,0^+$ results from the limit at $\infty$, setting $x=\dfrac 1t,\ t\to+\infty$.
