limit to infinity

**ryan18** · February 9th 2010, 08:46 AM

Could someone please show me how to workout problems of these types:

Find the limit. (If you need to use -

or

, enter -INFINITY or INFINITY.)

Thanks in advance!

**felper** · February 9th 2010, 08:51 AM

Originally Posted by ryan18

Could someone please show me how to workout problems of these types:

Find the limit. (If you need to use -

or

, enter -INFINITY or INFINITY.)

Thanks in advance!

Try multiplying by $\frac{\sqrt{36x^2+x}+6x}{\sqrt{36x^2+x}+6x}$ and then dividing by x up and down (i think that it have to works)

**ryan18** · February 9th 2010, 09:00 AM

How did you get the $\frac{\sqrt{36x^2+x}+36x^2}{\sqrt{36x^2+x}+36x^2}$ ?

**felper** · February 9th 2010, 09:15 AM

Originally Posted by ryan18

How did you get the $\frac{\sqrt{36x^2+x}+36x^2}{\sqrt{36x^2+x}+36x^2}$ ?

Oh, i did a little mistake. I've corrected it. Now that it's correct, with that factor, you can eliminate the root. See what happens:

$\lim_{x\to\infty}\sqrt{36x^2+x}-6x \cdot \frac{\sqrt{36x^2+x}+6x}{\sqrt{36x^2+x}+6x}=\lim_{ x\to\infty}\frac{x}{\sqrt{36x^2+x}+6x}=\lim_{x\to\ infty}\frac{1}{\sqrt{36+\frac{1}{x}}+6}$

Now you can conclude.

**ryan18** · February 9th 2010, 09:23 AM

How did you go from $<br /> =\lim_{x\to\infty}\frac{x}{\sqrt{36x^2+x}+6x}=\lim _{x\to\infty}\frac{1}{\sqrt{36+\frac{1}{x}}+6}<br />$
[/tex]?? It looks like in some places you multiplied out by $\frac{1}{x}$ and other parts you multiplied out by $\frac{1}{x^2}$

**felper** · February 9th 2010, 09:29 AM

Originally Posted by ryan18

How did you go from $<br /> =\lim_{x\to\infty}\frac{x}{\sqrt{36x^2+x}+6x}=\lim _{x\to\infty}\frac{1}{\sqrt{36+\frac{1}{x}}+6}<br />$
[/tex]?? It looks like in some places you multiplied out by $\frac{1}{x}$ and other parts you multiplied out by $\frac{1}{x^2}$

I've multiplied by 1/x both sides. $\frac{1}{x}(\sqrt{36x^2+x}+6x)=\frac{1}{x}(\sqrt{3 6x^2+x})+6=\frac{1}{\sqrt{x^2}}(\sqrt{36x^2+x})+6= \sqrt{36+\frac{1}{x}}$

**ryan18** · February 9th 2010, 09:36 AM

Ok I see now, so the denominator would look like $\sqrt{36+\frac{1}{x}}+6$ correct? and the $\frac{1}{x}$ would tend to 0. Leaving $\sqrt{36}+6=12$ ? Making the final answer to be $\frac{1}{12}$ ??

**felper** · February 9th 2010, 09:43 AM

Yep

**ryan18** · February 9th 2010, 09:46 AM

Awesome thank you so much for putting up with my multiple questions!

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