# Finding Limit as X approaches INFINITY

• February 6th 2008, 08:08 PM
topher0805
Finding Limit as X approaches INFINITY
Find the Limit:

http://www.webassign.net/www22/symIm...af64c9b82e.gif

I multiplied the numerator and denominator by the denominator, then started factoring things out and I ended up getting 1/64.

WebAssign is telling me this is wrong, but I can't figure out why. Any help?
• February 6th 2008, 08:30 PM
mr fantastic
Quote:

Originally Posted by topher0805
Find the Limit:

http://www.webassign.net/www22/symIm...731c0f0bae.gif

I multiplied the numerator and denominator by the denominator, then started factoring things out and I ended up getting 1/64.

WebAssign is telling me this is wrong, but I can't figure out why. Any help?

Things can get tricky with the $-\infty$. To avoid trouble, make the substitution t = -x. Then you want

$\lim_{t \rightarrow +\infty} (-t + \sqrt{t^2 - 2t}) = \lim_{t \rightarrow +\infty} (\sqrt{t^2 - 2t} - t)$.

Note that:

$\displaystyle \frac{(\sqrt{t^2 - 2t} - t)(\sqrt{t^2 - 2t} + t)}{\sqrt{t^2 - 2t} + t} = \frac{(t^2 - 2t) - t^2}{\sqrt{t^2 - 2t} + t} = \frac{-2t}{\sqrt{t^2 - 2t} + t} = \frac{-2}{\sqrt{1 - \frac{2}{t}} + 1}$.

$\lim_{t \rightarrow +\infty} \frac{-2}{\sqrt{1 - \frac{2}{t}} + 1} = \frac{-2}{1 + 1} = -1$.
• February 6th 2008, 08:34 PM
topher0805
Well, I did need help with that one too but the one I was actually describing is now in my original post.
• February 6th 2008, 08:49 PM
mr fantastic
Quote:

Originally Posted by topher0805
Well, I did need help with that one too but the one I was actually describing is now in my original post.

You can't just do an edit and completely change the question that's been replied to! It's poor form. Good job I quoted the original original post.

New question - new post.
• February 6th 2008, 08:50 PM
mr fantastic
Quote:

Originally Posted by topher0805
Find the Limit:

http://www.webassign.net/www22/symIm...af64c9b82e.gif

I multiplied the numerator and denominator by the denominator, then started factoring things out and I ended up getting 1/64.

WebAssign is telling me this is wrong, but I can't figure out why. Any help?

Divide top and bottom by x and then take the limit:

$\lim_{x \rightarrow \infty}\frac{x + 2}{\sqrt{64x^2 + 1}} = \lim_{x \rightarrow \infty}\frac{1 + \frac{2}{x}}{\sqrt{64 + \frac{1}{x^2}}} = \frac{1 + 0}{\sqrt{64 + 0}} = \frac{1}{8}$