why didn't i solve the lim well?

• Apr 25th 2013, 11:15 PM
orir
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}
$
• Apr 25th 2013, 11:25 PM
ibdutt
Re: why didn't i solve the lim well?
I find the approach absolutely correct.
• Apr 25th 2013, 11:29 PM
MarkFL
Re: why didn't i solve the lim well?
I don't see where you went wrong myself, and your result agrees with mine.
• Apr 26th 2013, 12:10 AM
orir
Re: why didn't i solve the lim well?
• Apr 26th 2013, 12:27 AM
MarkFL
Re: why didn't i solve the lim well?
Yes, and as $x\to\infty$ the limit is 1/2.
• Apr 26th 2013, 01:02 AM
orir
Re: why didn't i solve the lim well?
Quote:

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?
• Apr 26th 2013, 01:12 AM
MarkFL
Re: why didn't i solve the lim well?
I missed the negative sign...in that case I would rewrite the limit as:

$\lim_{x\to\infty}\left(\sqrt{x^2-x}+x \right)$
• Apr 26th 2013, 02:11 AM
orir
Re: why didn't i solve the lim well?
didn't understand you.. what in this case?
• Apr 26th 2013, 02:29 AM
MarkFL
Re: why didn't i solve the lim well?
In the case of $x\to-\infty$ I substituted x with -x...
• Apr 26th 2013, 03:52 AM
zzephod
Re: why didn't i solve the lim well?
Quote:

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.

.
• Apr 26th 2013, 04:17 AM
orir
Re: why didn't i solve the lim well?
ok.. so i still don't get how it should be solved
• Apr 26th 2013, 05:23 AM
Plato
Re: why didn't i solve the lim well?
Quote:

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.