It looks messy, but also familiar

Note that $(\sqrt{m+1}+\sqrt{m})^{-1} = \sqrt{m+1}-\sqrt{m}$ (this can be proved using the identity $a^2-b^2 = (a+b)(a-b)$ , replacing $a$ with $\sqrt{m+1}$ and $b$ with $\sqrt{m}$ )

When we plug in the summation, we will get $\displaystyle \sum_{m=1}^{n^2-1}(\sqrt{m+1}-\sqrt{m})$ , which equals to $(\sqrt{n^2}-\sqrt{n^2-1})+(\sqrt{n^2-1}-\sqrt{n^2-2})+\dots+(\sqrt{2}-\sqrt{1}) = \sqrt{n^2}-1=|n|-1$ However, given $n$ is a positive integer, we can safely remove the absolute value. Upon solving the equation completely, it will give $\log_{n}n = 1$

Notice that we don't touch the limit at all as all we should do is to simplify the inside first. When we are done, it gives integer, and the limit to any point of integer function is that integer itself.

Well $\ddot\smile$