Limit of function $f(x) = \sqrt{(xa + d)^2 + x^2 b^2} - \sqrt{(xa - d)^2 + x^2 b^2}$ 
I tried to calculate limit when $x$ goes to infinity for the following function 
  $$f(x) = \sqrt{(xa + d)^2 + x^2 b^2} - \sqrt{(xa - d)^2 + x^2 b^2}$$
  where $a$, $b$, $d$  are some positive constants.

It's easy to see that terms before and after minus sign goes to infinity so that gives me indeterminate symbol. Is there some way to solve this problem?
 A: Multiplying by $\sqrt{\cdot} + \sqrt{\cdot}$ at numerator and denomitor your get
$$\begin{aligned}f(x)&= \frac{4adx}{{\sqrt{(xa + d)^2 + x^2 b^2} + \sqrt{(xa - d)^2 + x^2 b^2}}}\\
\end{aligned}$$
And therefore $\lim\limits_{x \to \infty} f(x)  = \frac{2ad}{\sqrt{a^2+b^2}}$ by pulling $x$ at the denominator as
$$\begin{aligned}\sqrt{(xa + d)^2 + x^2 b^2} + \sqrt{(xa - d)^2 + x^2 b^2}&=x\left(\sqrt{(a+ d/x)^2 + b^2} + \sqrt{(a - d/x)^2 + b^2}\right)\\
\end{aligned}$$ for $x>0$ and $\lim\limits_{x \to \infty} d/x =0$.
Easy then to get $\lim\limits_{x \to -\infty} f(x)$ as $f$ is odd.
A: The simplest solution is probably the one in mathcounterexamples.net's answer.
But that's just an algebraic trick. You can also give a solution using calculus:


If $0<s<t$ there exists $\xi\in(s,t)$ such that $\sqrt t-\sqrt s=\frac{t-s}{2\sqrt\xi}$.


Proof: Apply the Mean Value Theorem to the function $f(x)=\sqrt x$.
Now let $t = (xa+d)^2+x^2b^2$ and $s=(xa-d)^2+x^2b^2$ and see what happens...
(One reason to do this is that this method can be applied in other situations where the algebraic trick is not available. MVT is the source of all wisdom.)
Details added in response to a comment: Let's say $a$, $x$ annd $d$ are positive, so if we  define $s$ and $t$ as above we have $s<t$. Now $s<\xi<t$ leads to $$\frac{t-s}{2\sqrt  t}<\frac{t-s}{2\sqrt \xi}<\frac{t-s}{2\sqrt s},$$or in other words $$\frac{2xad}{\sqrt{(xa+d)^2+x^2b^2}}<\frac{t-s}{2\sqrt\xi}<\frac{2xad}{\sqrt{(xa-d)^2+x^2b^2}},$$or$$\frac{2ad}{\sqrt{(a+d/x)^2+b^2}}<\frac{t-s}{2\sqrt\xi}<\frac{2ad}{\sqrt{(a-d/x)^2+b^2}}.$$Now let $x\to\infty$ and you're done.
(Hehe: When I posted this answer I hadn't worked out those details; it just seemed clear that  MVT "must" tell us what we need to know about $f(t)-f(s)$. Sure enough, MVT is the key to understanding the universe.)
The Moral
Any time you need to figure out something about a quantity of the form $f(t)-f(s)$ (where $f$ is differentiable) you should always consider applying MVT. Because MVT always tells you something about $f(t)-f(s)$, and if $t-s$ is not too big that "something"  will often include everything there is to know about $f(t)-f(s)$. (Not going to give a precise definition of "not too big" here - work out what MVT tells you, and if that's what you needed to know then $t-s$ was not  too big...)
A: For $x\not=0$,
$$\begin{align}\sqrt{(xa \pm d)^2 + x^2 b^2} &=|x|\sqrt{\left(a \pm \frac{d}{x}\right)^2+ b^2}\\
&=|x|\sqrt{a^2+b^2}\left(1\pm\frac{2ad}{(a^2+b^2)x}+o(1/x)\right)^{1/2}\\
&=|x|\sqrt{a^2+b^2}\left(1\pm\frac{ad}{(a^2+b^2)x}+o(1/x)\right)\\
&=|x|\sqrt{a^2+b^2}\pm\frac{ad|x|/x}{\sqrt{a^2+b^2}}+o(1)
\end{align}$$
where we used the expansion at $t=0$, $(1+t)^{1/2}=1+t/2+o(t)$.
Now it should be easy to verify that
$$\lim\limits_{x \to +\infty} f(x)  = \frac{2ad}{\sqrt{a^2+b^2}}\quad\text{and}\quad \lim\limits_{x \to -\infty} f(x)  = -\frac{2ad}{\sqrt{a^2+b^2}}.$$
A: Let $t=1/(2 d x)$. Then the limit of interest can be written as 
$$\lim_{t\to 0^{\pm}}\left(\sqrt{(2a d/t + d)^2 + b^2 d^2/t^2} - \sqrt{(2a d/t - d)^2 + b^2 d^2/t^2}\right)\\
\hspace{4cm}=2d\,\text{sgn}(t)\times \lim_{t\to 0^{\pm}}\frac{1}{t}\left[\sqrt{(a+t/2)^2+b^2}-\sqrt{(a-t/2)^2+b^2}\right].$$
The definition of the derivative therefore yields the limits for $t\to 0^{\pm}$ as
$$\pm 2d \frac{d}{dt}\left( \sqrt{(a+t)^2+b^2}\right)_{t=0}=\pm \frac{2ad}{\sqrt{a^2+b^2}}.$$
