Evaluating the limit: $\lim_{x\rightarrow -1^+}\sqrt[3]{x+1}\ln(x+1)$

Question

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.

Answer 1

$$\lim_{x\rightarrow -1^+}\sqrt[3]{x+1}\ln(x+1) = \lim_{x\rightarrow 0^+}\sqrt[3]{x}\ln(x) = \lim_{x\rightarrow 0^+} e^{ln (x)\frac{1}{3}}\ln(x) = \lim_{x\rightarrow -\infty} e^{\frac{1}{3}x} x$$

$$= -\lim_{x\rightarrow +\infty} \frac{x}{e^{\frac{1}{3}x}} = -\lim_{x\rightarrow +\infty} \frac{x}{1 + \frac{x}{3} + \frac{x^2}{18} +O(x^3)} = -\lim_{x\rightarrow +\infty} \frac{1}{\frac{1}{x} + \frac{1}{3} + \frac{x}{18} +O(x^3)} = 0$$

Answer 2

If you set $u:=(x+1)$, then we are looking for $$ \lim_{u \to 0^+} u^{1/3} \ln u=3\lim_{v \to 0^+} v \ln v. $$ Now a proof (without l'Hospital rule) of

$$ \lim_{v \to 0^+} v \ln v=0 $$

can be found for example here. Thus your initial limit is equal to $0$.

Answer 3

$$\lim\limits_{x\to -1^+}\sqrt[3]{x+1}\ln(x+1)$$ Let $h=x+1$. Since $x\to -1^+$, then $h\to 0^+$. So now $$\lim\limits_{h\to 0^+}\sqrt[3]{h}\ln h$$ $$=\lim\limits_{h\to 0^+}h^{\frac13}\ln h$$ $$=\lim\limits_{h\to 0^+}\exp\left(\ln h^{\frac13}\right)\ln h$$ $$=\lim\limits_{h\to 0^+}\exp\left(\frac13\ln h\right)\ln h$$ $$=3\lim\limits_{h\to 0^+}\left(\frac13\ln h\right)\exp\left(\frac13\ln h\right)$$ Let $k=\frac13\ln h$. Since $h\to 0^+$, then $k\to -\infty$. So now $$3\lim\limits_{k\to -\infty}ke^k$$ $$=3\lim\limits_{k\to -\infty}\frac{k}{e^{-k}}$$ Let $m=-k$. Since $k\to -\infty$, then $m\to\infty$. So now $$-3\lim\limits_{m\to \infty}\frac{m}{e^m}=0$$ At this point it should be clear that this limit is zero.

Answer 4

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 $x=\dfrac 1t,\ t\to+\infty$.