How do I calculate $\lim_{x\to+\infty}\sqrt{x+a}-\sqrt{x}$? I've seen a handful of exercises like this:
$$\lim_{x\to+\infty}(\sqrt{x+a}-\sqrt{x})$$
I've never worked with limits to infinity when there is some arbitrary number $a$. I am not given any details about it.
Apparently the answer is $0$. How was that conclusion reached?

My guess is that since $x = +\infty$, the result of $x + a$ will still be $+\infty$ so we would have $\sqrt{x}-\sqrt{x} = 0$.
But that doesn't convince me. For starters,  we don't know what $a$ is: it could be $-\infty$ or something, so $\infty - \infty$ would be indeterminate...
 A: When an expression like that is written, the common assumption (in a Calculus course) is that $a$ is some real number. 
The symbols $\infty$ of $-\infty$ do not represent real numbers and cannot be treated as such. Expressions like $x\to\infty$ and $\lim f(x)=\infty$ have precise meanings that do not involve the symbol $\infty$. 
In this case,
$$
\sqrt{x+a}-\sqrt x=(\sqrt{x+a}-\sqrt x)\,\frac{\sqrt{x+a}+\sqrt x}{\sqrt{x+a}+\sqrt x}
=\frac{a}{\sqrt{x+a}+\sqrt x}.
$$
For whatever number $a$ is, the expression on the right goes to zero as $x\to\infty$. To argue formally, note that 
$$
\left|\frac{a}{\sqrt{x+a}+\sqrt x}\right|=\frac{|a|}{\sqrt{x+a}+\sqrt x}\leq\frac{|a|}{\sqrt x}.
$$
For a given $\varepsilon>0$ (that we think as small), if $x>a^2/\varepsilon^2$, then 
$$
\frac{|a|}{\sqrt x}<\frac{|a|}{|a|/\varepsilon}=\varepsilon.
$$
This means that, if $x$ is big enough we can guarantee that $\frac{a}{\sqrt{x+a}+\sqrt x}$ it is as close to zero as we want. 
A: METHOD 1:
We can write
$$\begin{align}
\sqrt{x+a}&=\sqrt{x}\left(1+\frac ax\right)^{1/2}\\\\
&=\sqrt{x}\left(1+\frac{a}{2x}+O\left(\frac ax\right)^2\right)\\\\
&=\sqrt{x}+\frac{a}{2\sqrt{x}}+O\left(\frac 1x\right)^{3/2}
\end{align}$$
Thus, 
$$\sqrt{x+a}-\sqrt{x}=\frac{a}{2\sqrt{x}}+O\left(\frac 1x\right)^{3/2}\to 0$$
as $x\to \infty$.

METHOD 2:
$$\begin{align}
\sqrt{x+a}-\sqrt{x}&=\left(\sqrt{x+a}-\sqrt{x}\right)\,\frac{\sqrt{x+a}+\sqrt{x}}{\sqrt{x+a}+\sqrt{x}}\\\\
&=\frac{a}{\sqrt{x+a}+\sqrt{x}}\to 0
\end{align}$$
as $x\to \infty$.
A: Notice, $$\lim_{x\to \infty}(\sqrt{x+a}-\sqrt x)$$
$$=\lim_{x\to \infty}(\sqrt{x+a}-\sqrt x)\frac{(\sqrt{x+a}+\sqrt x)}{(\sqrt{x+a}+\sqrt x)}$$
$$=\lim_{x\to \infty}\frac{x+a-x}{\sqrt{x+a}+\sqrt x}$$
$$=a\lim_{x\to \infty}\frac{1}{\sqrt{x+a}+\sqrt x}=a(0)=0$$
A: You may write
$$
\sqrt{x+a}-\sqrt{x}=\left(\sqrt{x+a}-\sqrt{x}\right)\frac{\sqrt{x+a} + \sqrt{x}}{\sqrt{x+a} + \sqrt{x}}=\frac{a}{\sqrt{x+a} + \sqrt{x}}
$$ then it becomes easier to obtain your limit as $x \to \infty$.
