Limit of product of mixed root indexes

Question

I am unable to recover the limit of this function.

$$ \lim_{x \to +\infty} \left (\frac{1}{\sqrt{x}} \times \sqrt[3]{x+1} \right ) $$

I have tried many ways of solving it. These are the two simplest:

• Variable substitution

Using $y = \sqrt[3]{x+1} $, with which I got a common factor $\sqrt{y-1}$ and got

$$ \lim_{x \to +\infty} \left (\frac{1}{\sqrt{y^3-1}} \times y \right) = \lim_{x \to +\infty} \left (\frac{\sqrt{y^2}}{\sqrt{y^3-1}} \right) = \lim_{x \to +\infty} \left (\frac{\sqrt{y^2}}{\sqrt{y^3(1-\frac{1}{y^3})}} \right) = \lim_{x \to +\infty} \left (\frac{1}{\sqrt{y \left (1-\frac{1}{y^3} \right)}} \right) = \frac{1}{+\infty \times 1} = 0 $$

• Using the common factor $\sqrt{x}$

$\sqrt[3]{x+1} = \sqrt[6]{(x+1)^2} = \sqrt[6]{x^2+2x+1} = \sqrt[6]{x^3 (\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x^3})}=\sqrt{x} \left (\sqrt[6]{\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x^3}} \right )$

and therefore I got

$$ \lim_{x \to +\infty} \left (\frac{\sqrt{x} \left (\sqrt[6]{\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x^3}} \right )}{\sqrt{x}} \right ) = \lim_{x \to +\infty} \left (\sqrt[6]{\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x^3}} \right ) = \sqrt[6]{\frac{1}{+\infty}+\frac{1}{+\infty}+\frac{1}{+\infty}} = 0 $$

However, I found that the limit is positive when I plotted a function with this expression. I have plotted it here, you can look at it yourselves. This means my calculations are wrong. What am I doing wrong and how do I solve it correcly? Thank you for the help.

Answer 1

Your calculations are right! $$ \begin{aligned} \lim _{x\to \infty }\left(\frac{\sqrt[3]{1+x}}{\sqrt{x}}\right) & = \lim _{t\to 0\:}\left(\frac{\sqrt[3]{1+\frac{1}{t}}}{\sqrt{\frac{1}{t}}}\right) \\& = \lim _{t\to 0\:}\left(\sqrt[6]{t}\sqrt[3]{t+1}\right) \\& =0^{\frac{1}{6}}\sqrt[3]{0+1} = \color{red}{0} \end{aligned} $$

Answer 2

Multiplying and dividing by $x^{1/3}$ we get $\frac{1}{\sqrt{x}}(x+1)^{1/3}=\frac{x^{1/3}}{\sqrt{x}}(\frac{x+1}{x})^{1/3}$. The factor $(\frac{x+1}{x})^{1/3}$ tends to $1$ as $x \to \infty$. The factor $\frac{x^{1/3}}{\sqrt{x}}$ equals $x^{-1/6}$ that tends to $0$ as $x \to \infty$. So the limit you're looking for is $0$.