limit of $\sqrt{(x^2+1)/(x^3+1)}$ as $x$ approaches negative infinity

Question

$\lim_{x \rightarrow - \infty } \sqrt{ \frac{x^2+1}{x^3+1} } $ My teacher says that no limit exists, but Wolfram Alpha says the limit is 0. I'm confused. Any helps are welcome.

Answer 1

Let $-1/x=h\implies h\to0^+$

$$\implies\sqrt{\dfrac{1+x^2}{1+x^3}}=\sqrt{\dfrac{1+\dfrac1{h^2}}{1-\dfrac1{h^3}}}=\sqrt{\dfrac{h(h^2+1)}{h^3-1}}\text{ as }h\ne0$$

Answer 2

\begin{align} &\lim_{x \to -\infty} \sqrt{\frac{1+x^{2}}{1+x^{3}}} \\ = &\lim_{x \to -\infty} \sqrt{\frac{\frac{1}{x^{3}}+ \frac{1}{x}}{\frac{1}{x^{3}}+1}} \qquad \mathrm{divide \; numerator \; and \; denominator \; by \;} x^{3}\\ = &\frac{0 + 0}{0 + 1} \\ = &\quad 0 \end{align}

Answer 3

The answer depends on the underlying domain of the expression. To see this, you may write

$$\sqrt{ \frac{x^2+1}{x^3+1} } \stackrel{x<-1}{=} \sqrt{-1} \cdot \frac{1}{\sqrt{|x|}} \cdot \sqrt{ \frac{1+\frac{1}{x^2}}{1+\frac{1}{x^3}} } = \begin{cases} \mbox{not defined} & \mbox{ in } \mathbb{R} \\ \pm i \cdot \frac{1}{\sqrt{|x|}} \cdot \sqrt{ \frac{1+\frac{1}{x^2}}{1+\frac{1}{x^3}} } \stackrel{x \to -\infty}{\longrightarrow} 0 & \mbox{ in } \mathbb{C}\end{cases}$$