# Thread: Limit Problem

1. ## Limit Problem

Please help with this problem i have final tomorrow

2. ## Re: Limit Problem

Seems to tend to infinity, no?

3. ## Re: Limit Problem

Originally Posted by Telo
Please help with this problem i have final tomorrow

\displaystyle \displaystyle \begin{align*} \sqrt{n^2 + 3} + \sqrt{n^2 + 8} &> \sqrt{n^2} + \sqrt{n^2} \\ &= |n| + |n| \\ &= 2|n| \end{align*}

Since \displaystyle \displaystyle \begin{align*} 2|n| \to \infty \end{align*}, therefore so must \displaystyle \displaystyle \begin{align*} \sqrt{n^2 + 3} + \sqrt{n^2+8} \end{align*}.

4. ## Re: Limit Problem

Originally Posted by Telo
Please help with this problem i have final tomorrow

Quite frankly this is a trivial problem. As the others have said, as n goes to infinity, both parts do also and so their sum does.

A much more interesting problem would be $\displaystyle \lim_{x\to\infty}\left[\sqrt{n^2+ 3}- \sqrt{n^2+ 8}\right]$. For that, I would recommend multiplying by the "unit fraction" $\displaystyle \frac{\sqrt{n^2+ 3}+ \sqrt{n^2+ 8}}{\sqrt{n^2+3}+ \sqrt{n^2+ 8}}$.

5. ## Re: Limit Problem

Originally Posted by Telo
Please help with this problem i have final tomorrow
You have a final the day after Christmas?