# Thread: Limit to infinity question

1. ## Limit to infinity question

I'm really having trouble with this question...
When I solve it I find the answer to just be zero but apparently the answer is two or -2.

Find lim $1/(((x^2+x)^1/2)+x)$
x-> - infinity

Could somebody please explain to me how to do this type of question. I understand what to do when there are variables on the top, but since there is only a one I'm very confused.

2. ## Re: Limit to infinity question

Originally Posted by poipoipoi10
When I solve it I find the answer to just be infinity but apparently the answer is two or -2.
Find lim 1/(((x^2+x)^1/2)+x)
x-> infinity
If the question is $\lim _{x \to \infty } \frac{1}{{\sqrt {x^2 + x} + x}}$ then the limit is 0.

3. ## Re: Limit to infinity question

Oh man, sorry I meant to type negative infinity. But wouldnt that still make it 0?

4. ## Re: Limit to infinity question

Originally Posted by poipoipoi10
Oh man, sorry I meant to type negative infinity. But wouldnt that still make it 0?
No, that makes a real difference.
In that case because if $x<0$ then $\frac{1}{{\sqrt {x^2 + x} + x}} = - \sqrt {1 + \frac{1}{x}} - 1$

5. ## Re: Limit to infinity question

Would that not still make the numerator 0? Because How I learned to do it was divide by the highest power denominator, which in this case would be $x^2$

6. ## Re: Limit to infinity question

Originally Posted by poipoipoi10
Would that not still make the numerator 0? Because How I learned to do it was divide by the highest power denominator, which in this case would be $x^2$
The answer is $-2$.
The working is just algebra. And using $|x|=\sqrt{x^2}$ because $x<0$.

7. ## Re: Limit to infinity question

I'm fairly new to this website so I dont know how to format all my algebra the way that you guys do =(. But this is my train of thought when I do the question:

1. Divide by highest power x in denominator.

(1/x^2)/-(((1/x^4)^1/2)((x^2+x)^1/2)+x/x^2)

2. Continue algebra (divide into radical).

(1/x^2)/-(((x^2/x^4 + x/x^4)^1/2) + x/x^2)

3. Simplify

(1/x^2)/-(((1/x^2 + 1/x^3)^1/2) + 1/x)

4. Plug infinity's in

0/-((0+0)^2)+0)

So basically I end up with 0/0.

I'm sorry that it's so hard to understand, I wish I could format it the way you do, but I can't figure it out.

8. ## Re: Limit to infinity question

Originally Posted by poipoipoi10
1. Divide by highest power x in denominator.
Forget about dividing. We don't use it until the last step.

$\begin{gathered} \frac{1}{{\sqrt {x^2 + x} + x}}\frac{{\sqrt {x^2 + x} - x}}{{\sqrt {x^2 + x} - x}} \hfill \\ = \frac{{\sqrt {x^2 + x} - x}}{x} \hfill \\ = \frac{{\left| x \right|\sqrt {1 + \frac{1}{x}} - x}}{x} ,~~x<0\hfill \\ = - \sqrt {1 + \frac{1}{x}} - 1 \hfill \\ \end{gathered}$

9. ## Re: Limit to infinity question

Ah, that cleared it up for me, thanks =)

If you can still see this I have one quick question:

How in your third step did you get the 1/x when you took out the x^2?

10. ## Re: Limit to infinity question

Originally Posted by poipoipoi10
Ah, that cleared it up for me, thanks =)

If you can still see this I have one quick question:

How in your third step did you get the 1/x when you took out the x^2?
First, since you are making $\displaystyle x \to -\infty$, you need to know that $\displaystyle \sqrt{x^2} = -x$ if $\displaystyle x < 0$. So

\displaystyle \begin{align*} \sqrt{x^2 + x} &= \sqrt{x^2\left(1 + \frac{1}{x}\right)} \\ &= \sqrt{x^2}\sqrt{1 + \frac{1}{x}} \\ &= -x\sqrt{1 + \frac{1}{x}} \end{align*}