1. ## limit to infinity

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.)

2. 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.)

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

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

4. Originally Posted by ryan18
How did you get the $\displaystyle \frac{\sqrt{36x^2+x}+36x^2}{\sqrt{36x^2+x}+36x^2}$?
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:

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

Now you can conclude.

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

6. Originally Posted by ryan18
How did you go from $\displaystyle =\lim_{x\to\infty}\frac{x}{\sqrt{36x^2+x}+6x}=\lim _{x\to\infty}\frac{1}{\sqrt{36+\frac{1}{x}}+6}$
[/tex]?? It looks like in some places you multiplied out by $\displaystyle \frac{1}{x}$ and other parts you multiplied out by $\displaystyle \frac{1}{x^2}$
I've multiplied by 1/x both sides. $\displaystyle \frac{1}{x}(\sqrt{36x^2+x}+6x)=\frac{1}{x}(\sqrt{3 6x^2+x})+6=\frac{1}{\sqrt{x^2}}(\sqrt{36x^2+x})+6= \sqrt{36+\frac{1}{x}}$

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

8. Yep

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