2. ## Re: Limit Problem

Seems to tend to infinity, no?

3. ## Re: Limit Problem

Originally Posted by Telo

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

4. ## Re: Limit Problem

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