<s>[INST] 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.
 [/INST] $$\lim_{x\rightarrow -1^+}\sqrt[3]{x+1}\ln(x+1) = \lim_{x\rightarrow 0^+}\sqrt[3]{x}\ln(x) = \lim_{x\rightarrow 0^+} e^{ln (x)\frac{1}{3}}\ln(x) = \lim_{x\rightarrow -\infty} e^{\frac{1}{3}x} x$$
$$= -\lim_{x\rightarrow +\infty}  \frac{x}{e^{\frac{1}{3}x}} = -\lim_{x\rightarrow +\infty}  \frac{x}{1 + \frac{x}{3} + \frac{x^2}{18} +O(x^3)} = -\lim_{x\rightarrow +\infty}  \frac{1}{\frac{1}{x} + \frac{1}{3} + \frac{x}{18} +O(x^3)} = 0$$
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