Finding the limit as x tends to infinity

**Glitch** · July 26th 2010, 05:30 AM

The question:

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

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:

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

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

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

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.

**Chris L T521** · July 26th 2010, 05:39 AM

Originally Posted by Glitch

The question:

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

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:

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

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

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

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 $\lim\limits_{x\to\infty}\dfrac{1}{\sqrt{1+\frac{1} {x}}+1}$

Can you solve it now?

**Glitch** · July 26th 2010, 05:44 AM

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:

**Danny** · July 26th 2010, 05:52 AM

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

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

**HallsofIvy** · July 26th 2010, 06:00 AM

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

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

**Glitch** · July 26th 2010, 06:30 AM

Ahh, thanks!

**Archie Meade** · July 26th 2010, 07:10 AM

Alternatively,

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

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

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

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

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