# Math Help - why didn't i solve the lim well?

1. ## why didn't i solve the lim well?

$\lim_{x\rightarrow-\infty}(\sqrt{x^{2}+x}-x) = ?$. this is how i solve this, but i know it's not true.. why? $
(\sqrt{x^{2}+x}-x)=\frac{(\sqrt{x^{2}+x}-x)\cdot(\sqrt{x^{2}+x}+x)}{(\sqrt{x^{2}+x}+x)}=$
$\frac{x}{\sqrt{x^{2}+x}+x}$
$\frac{x}{x+x\cdot\sqrt{1+\frac{1}{x}}}=\frac{1}{1+ \sqrt{1+\frac{1}{x}}}$ so, $\lim_{x\rightarrow-\infty}f(x)=\frac{1}{2}
$

2. ## Re: why didn't i solve the lim well?

I find the approach absolutely correct.

3. ## Re: why didn't i solve the lim well?

I don't see where you went wrong myself, and your result agrees with mine.

5. ## Re: why didn't i solve the lim well?

Yes, and as $x\to\infty$ the limit is 1/2.

6. ## Re: why didn't i solve the lim well?

Originally Posted by MarkFL
Yes, and as $x\to\infty$ the limit is 1/2.
and what about the limit as $x\rightarrow-\infty$ , which it's the case we're dealing with?

7. ## Re: why didn't i solve the lim well?

I missed the negative sign...in that case I would rewrite the limit as:

$\lim_{x\to\infty}\left(\sqrt{x^2-x}+x \right)$

8. ## Re: why didn't i solve the lim well?

didn't understand you.. what in this case?

9. ## Re: why didn't i solve the lim well?

In the case of $x\to-\infty$ I substituted x with -x...

10. ## Re: why didn't i solve the lim well?

Originally Posted by orir
$\lim_{x\rightarrow-\infty}(\sqrt{x^{2}+x}-x) = ?$. this is how i solve this, but i know it's not true.. why? $(\sqrt{x^{2}+x}-x)=\frac{(\sqrt{x^{2}+x}-x)\cdot(\sqrt{x^{2}+x}+x)}{(\sqrt{x^{2}+x}+x)}=$ $\frac{x}{\sqrt{x^{2}+x}+x}$
$\frac{x}{x+x\cdot\sqrt{1+\frac{1}{x}}}=\frac{1}{1+ \sqrt{1+\frac{1}{x}}}$ so, $\lim_{x\rightarrow-\infty}f(x)=\frac{1}{2}$
Because at the begining of the last line you should have had:

$\frac{x}{x+|x|\cdot\sqrt{1+\frac{1}{x}}}$

rather than:

$\frac{x}{x+x\cdot\sqrt{1+\frac{1}{x}}}$

If you want to try going down that route.

.

11. ## Re: why didn't i solve the lim well?

ok.. so i still don't get how it should be solved

12. ## Re: why didn't i solve the lim well?

Originally Posted by orir
i still don't get how it should be solved

Please note the problem is $\lim_{x\rightarrow{\color{red}-}\infty}(\sqrt{x^{2}+x}-x) = ?$

In effect it is $\infty{\color{blue}+}\infty$.

To see that let $x=-10^6$ then we have $\sqrt{10^{12}-10^6}+10^6=10^3\sqrt{10^2-1}+10^6$, which is large.