the limit $\lim_{a \to \infty}\int_{a}^{a+1}\frac{x+\sqrt{x}}{x-\sqrt{x}}\,\mathrm dx=1$ I proved this using mean value theorem
$$\lim_{a \to \infty}\int_{a}^{a+1}\frac{x+\sqrt{x}}{x-\sqrt{x}}\,\mathrm dx=1$$ 
because $$\frac{x+\sqrt{x}}{x-\sqrt{x}}-1=\frac{2\sqrt{x}}{x-\sqrt{x}}$$
it suffices prove that $$\int_{a}^{a+1}\frac{2\sqrt{x}}{x-\sqrt{x}}\,\mathrm dx\rightarrow0  \text{ as} \ a\to \infty $$
the mean value theorem applies in the interval $[a,a+1]$.
My question is: Can I prove this with another argument?
My idea was this : $$\int_{a}^{a+1}\frac{2\sqrt{x}}{x-\sqrt{x}}dx=\int_{a}^{a+1}\frac{2}{\sqrt{x}-1}dx=\frac{2}{\sqrt{c}-1}$$ for some $c\in [a,a+1]$

Similar idea: Suppose $x>0$ and let $f(x) = \frac{x+\sqrt{x}}{x-\sqrt{x}}= \frac{1+\frac{1}{\sqrt{x}}}{1-\frac{1}{\sqrt{x}}}$. It is clear that $\lim_{x \to \infty} f(x) = 1$.
Let $\epsilon>0$ and choose $N$ large enough so that if $x \ge N$, then $|f(x)-1| < \epsilon$. Hence if $a > N$, we have 
$ |\int_a^{a+1} ( f(x) - 1)dx| \le \int_a^{a+1} | f(x) - 1|dx < \epsilon $.
Alternatively:
Note that if $x>1$, we have $f(x) = (1+\frac{1}{\sqrt{x}})(1+\frac{1}{(\sqrt{x})^1}+\frac{1}{(\sqrt{x})^2}+\cdots)$, or,
$f(x) = 1 + 2(\frac{1}{(\sqrt{x})^1}+\frac{1}{(\sqrt{x})^2}+\cdots)$, and furthermore, for any $N>1$, the convergence is uniform for $x \ge N$. Hence, for $a >1$, we can form the estimate
\begin{eqnarray}
|\int_a^{a+1} f(x) dx -1| &=& 2\int_a^{a+1} (\frac{1}{(\sqrt{x})^1}+\frac{1}{(\sqrt{x})^2}+\cdots) dx  \\
&\le& 2 \sum_{n=1}^\infty \frac{1}{(\sqrt{a})^n} \\
&=& 2 \frac{1}{1-\frac{1}{\sqrt{a}}}
\end{eqnarray}
The desired result follows.
