# Limit of product of mixed root indexes

I am unable to recover the limit of this function.

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

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

• Variable substitution

Using $y = \sqrt{x+1}$, 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 $\sqrt{x}$

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

and therefore I got

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

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.

• The link does not work – vrugtehagel Apr 23 '17 at 21:08
• $\lim_{x \to +\infty} \left (\frac{1}{\sqrt{x}} \times \sqrt{x+1} \right )= \lim_{x \to +\infty} \left (\frac{1}{\sqrt{x^3}} \times \sqrt{(x+1)^2} \right )= \sqrt{ \lim_{x \to +\infty} \left (\frac{{(x+1)^2} }{{x^3}} \right ) }=0$ – N74 Apr 23 '17 at 22:04

Your calculations are right! \begin{aligned} \lim _{x\to \infty }\left(\frac{\sqrt{1+x}}{\sqrt{x}}\right) & = \lim _{t\to 0\:}\left(\frac{\sqrt{1+\frac{1}{t}}}{\sqrt{\frac{1}{t}}}\right) \\& = \lim _{t\to 0\:}\left(\sqrt{t}\sqrt{t+1}\right) \\& =0^{\frac{1}{6}}\sqrt{0+1} = \color{red}{0} \end{aligned} • Thank you very much for posting the plot as well! I had checked the link and I could not find any better way of posting it. If you could tell me how I can do this as you did and save you the work for the future, I would really appreciate further. And as you see, if the limit is zero as I also think, it goes to zero very, very slowly. Also, I had seen a book stating the limit was some $\frac{\sqrt{2}}{a}$ being this $a$ a number I don't remember. When I checked the plot I have also seen this odd plateauing so I got concerned. – BrunoCAfonso Apr 23 '17 at 23:37
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 $0$.