# Thread: Finding the limit as x tends to infinity

1. ## Finding the limit as x tends to infinity

The question:

Find the limit as x tends to infinity $\displaystyle \sqrt{x^2 + x} - x$

I'm not sure how to go about it. I'm thinking that I need to get rid of that square root, or at least, get it in a form which will make it easy to read off the limit.

I've tried the following:

$\displaystyle \frac{\sqrt{x^2 + x} - x}{1} . \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x}$

$\displaystyle \frac{x^2 + x - x^2}{\sqrt{x^2 + x} + x}$

$\displaystyle \frac{x}{\sqrt{x^2 + x} + x}$

I'm not sure how to factor out x in the denominator, I'm guessing that'd simplify it.

If anyone could help, or offer a different method of solving this, I'd really appreciate it.

2. Originally Posted by Glitch
The question:

Find the limit as x tends to infinity $\displaystyle \sqrt{x^2 + x} - x$

I'm not sure how to go about it. I'm thinking that I need to get rid of that square root, or at least, get it in a form which will make it easy to read off the limit.

I've tried the following:

$\displaystyle \frac{\sqrt{x^2 + x} - x}{1} . \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x}$

$\displaystyle \frac{x^2 + x - x^2}{\sqrt{x^2 + x} + x}$

$\displaystyle \frac{x}{\sqrt{x^2 + x} + x}$

I'm not sure how to factor out x in the denominator, I'm guessing that'd simplify it.

If anyone could help, or offer a different method of solving this, I'd really appreciate it.
Divide both numerator and denominator by x to get $\displaystyle \lim\limits_{x\to\infty}\dfrac{1}{\sqrt{1+\frac{1} {x}}+1}$

Can you solve it now?

3. Thanks.

I know this is probably obvious (my brain is fried at the moment), but how did you divide the denominator? The square root is doing my head in. D:

4. $\displaystyle \dfrac{x}{\sqrt{x^2+x} + x} = \dfrac{x}{\sqrt{x^2\left(1+\frac{1}{x}\right)}+x} } = \dfrac{x}{x\sqrt{1+\frac{1}{x}}+x} }$

Then cancel $\displaystyle x$ to give you what Chris gave.

5. Or, (slightly) different way to do the same thing-
$\displaystyle \frac{\sqrt{x^2+ x}+ x}{x}= \sqrt{\frac{x^2+ x}{x^2}}+ 1= \sqrt{1+ \frac{1}{x}}+ 1$

You see how the "x" outside the square root became "$\displaystyle x^2$" inside the square root. As long as x is positive (which it certainly is here as x is going to $\displaystyle \infty$), $\displaystyle x= \sqrt{x^2}$.

6. Ahh, thanks!

7. Alternatively,

$\displaystyle \left(x^2+x\right)^n=\left(x^2\right)^n+\binom{n}{ 1}\left(x^2\right)^{n-1}x+\binom{n}{2}\left(x^2\right)^{n-2}x^2+....$

For n=0.5, as x approaches infinity, all negative powers of x approach 0.
Then the expansion reduces to

$\displaystyle \left(x^2\right)^{0.5}+0.5\left(x^2\right)^{-0.5}x=x+0.5$

$\displaystyle \displaystyle\Rightarrow\ \lim_{x\rightarrow\infty}\left(\sqrt{x^2+x}-x\right)=x-x+0.5=0.5$