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}

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

I find the approach absolutely correct.

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

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

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

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

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

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?

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

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

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

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

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

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

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

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.

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

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

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.