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 $c\in [a,a+1]$
 A: Let define $$F(a)=\int_2^a \frac{x+\sqrt{x}}{x-\sqrt{x}} dx$$
for $a\ge 2$. Since $f(x)=\frac{x+\sqrt{x}}{x-\sqrt{x}}$ is continuous on $[2,\infty)$, we know $F$ is differentiable on $(2,\infty)$. And by mean value theorem we get
$$\int_a^{a+1} f(x)dx=F(a+1)-F(a)=f(c)$$
for some $c\in(a,a+1)$. If we take $a\to\infty$, then $c\to\infty$. And we know
$$\lim_{c\to\infty}f(c)=\lim_{c\to\infty}\frac{c+\sqrt{c}}{c-\sqrt{c}}=1$$
so $\lim_{a\to\infty} F(a+1)-F(a)=1$.
A: 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.
A: $$\lim_{a \to \infty}\int_{a}^{a+1}\frac{x+\sqrt{x}}{x-\sqrt{x}}\,\mathrm dx=\lim_{a \to \infty}\int_{a}^{a+1}1+\frac{2\sqrt{x}}{x-\sqrt{x}}\,\mathrm dx$$ $$=1+\lim_{a \to \infty}\int_{a}^{a+1}\frac{2}{\sqrt{x}-1}\,\mathrm dx$$ 
A: Using the MVT is good. We do the same thing using an explicit estimate.
Let $f(x)= \frac{x+\sqrt{x}}{x-\sqrt{x}}$.
One can use the same idea as yours, noting more or less as you did that 
$$f(x)=1+\frac{2\sqrt{x}}{x-\sqrt{x}}=1+\frac{2}{\sqrt{x}-1}.$$
If $x\ge 4$, we have $\sqrt{x}-1 \ge (1/2)\sqrt{x}$. Thus if $x\ge 4$ we have
$$1\le f(x)\le 1+\frac{4}{\sqrt{x}}.$$
So if $a\ge 4$, our integral $I$ satisfies the inequality
$$1\le I \le 1+\frac{4}{\sqrt{a}}.$$
Now Squeezing gives the desired result.
A: This method might look difficult, but it is in fact a very common powerful tool.
Let $$f(x)=\frac{x+\sqrt{x}}{x-\sqrt{x}}.$$
Note that $f$ is positive and hence $$\left(\min_{a\leq t\leq a+1}f(t)\right)\int_a^{a+1}dx\leq\int_a^{a+1}f(x)dx\leq\left(\max_{a\leq t\leq a+1}f(t)\right)\int_a^{a+1}dx$$
Next, $\int_a^{a+1}dx=1$. 
It remains to calculate the limits of 
$$\max_{a\leq t\leq a+1}f(t)\qquad\text{and}\qquad \min_{a\leq t\leq a+1}f(t)$$ 
as $a\to\infty$, which should not be that hard.
