# why didn't i solve the lim well?

• Apr 25th 2013, 10:15 PM
orir
why didn't i solve the lim well?
$\displaystyle \lim_{x\rightarrow-\infty}(\sqrt{x^{2}+x}-x) = ?$. this is how i solve this, but i know it's not true.. why? $\displaystyle (\sqrt{x^{2}+x}-x)=\frac{(\sqrt{x^{2}+x}-x)\cdot(\sqrt{x^{2}+x}+x)}{(\sqrt{x^{2}+x}+x)}=$$\displaystyle \frac{x}{\sqrt{x^{2}+x}+x} \displaystyle \frac{x}{x+x\cdot\sqrt{1+\frac{1}{x}}}=\frac{1}{1+ \sqrt{1+\frac{1}{x}}} so, \displaystyle \lim_{x\rightarrow-\infty}f(x)=\frac{1}{2} • Apr 25th 2013, 10:25 PM ibdutt Re: why didn't i solve the lim well? I find the approach absolutely correct. • Apr 25th 2013, 10:29 PM MarkFL Re: why didn't i solve the lim well? I don't see where you went wrong myself, and your result agrees with mine. • Apr 25th 2013, 11:10 PM orir Re: why didn't i solve the lim well? • Apr 25th 2013, 11:27 PM MarkFL Re: why didn't i solve the lim well? Yes, and as \displaystyle x\to\infty the limit is 1/2. • Apr 26th 2013, 12:02 AM orir Re: why didn't i solve the lim well? Quote: Originally Posted by MarkFL Yes, and as \displaystyle x\to\infty the limit is 1/2. and what about the limit as \displaystyle x\rightarrow-\infty , which it's the case we're dealing with? • Apr 26th 2013, 12:12 AM MarkFL Re: why didn't i solve the lim well? I missed the negative sign...in that case I would rewrite the limit as: \displaystyle \lim_{x\to\infty}\left(\sqrt{x^2-x}+x \right) • Apr 26th 2013, 01:11 AM orir Re: why didn't i solve the lim well? didn't understand you.. what in this case? • Apr 26th 2013, 01:29 AM MarkFL Re: why didn't i solve the lim well? In the case of \displaystyle x\to-\infty I substituted x with -x... • Apr 26th 2013, 02:52 AM zzephod Re: why didn't i solve the lim well? Quote: Originally Posted by orir \displaystyle \lim_{x\rightarrow-\infty}(\sqrt{x^{2}+x}-x) = ? . this is how i solve this, but i know it's not true.. why? \displaystyle (\sqrt{x^{2}+x}-x)=\frac{(\sqrt{x^{2}+x}-x)\cdot(\sqrt{x^{2}+x}+x)}{(\sqrt{x^{2}+x}+x)}=$$\displaystyle \frac{x}{\sqrt{x^{2}+x}+x}$
$\displaystyle \frac{x}{x+x\cdot\sqrt{1+\frac{1}{x}}}=\frac{1}{1+ \sqrt{1+\frac{1}{x}}}$ so, $\displaystyle \lim_{x\rightarrow-\infty}f(x)=\frac{1}{2}$

Because at the begining of the last line you should have had:

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

rather than:

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

If you want to try going down that route.

.
• Apr 26th 2013, 03:17 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, 04:23 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 $\displaystyle \lim_{x\rightarrow{\color{red}-}\infty}(\sqrt{x^{2}+x}-x) = ?$

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

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