Confusion in evaluating the limit $\lim_{x\to-\infty}\sqrt{x^2+ax}-\sqrt{x^2+bx}$ I was solving a question related to functions and i come across a limit which i cannot understand.The question is 
If $a$ and $b$ are positive real numbers such that $a-b=2,$ then find the smallest value of the constant $L$ for which $\sqrt{x^2+ax}-\sqrt{x^2+bx}<L$ for all $x>0$

First i found the domain of definition of the function in question $\sqrt{x^2+ax}-\sqrt{x^2+bx}$.The domain is $x\geq0 \cup x\leq -a$.
Then i found the horizontal asymptotes as $x\to \infty$.
$\lim_{x\to\infty}\sqrt{x^2+ax}-\sqrt{x^2+bx}=\lim_{x\to\infty}\frac{(a-b)x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\lim_{x\to\infty}\frac{2x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}$
$=\lim_{x\to\infty}\frac{2}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}=\frac{2}{2}=1$
Then i found the horizontal asymptotes as $x\to -\infty$.
$\lim_{x\to-\infty}\sqrt{x^2+ax}-\sqrt{x^2+bx}=\lim_{x\to-\infty}\frac{(a-b)x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\lim_{x\to-\infty}\frac{2x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}$
$=\lim_{x\to-\infty}\frac{2}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}=\frac{2}{2}=1$ 
But when i drew the graph using graphing calculator,the horizontal asymptote as $x\to-\infty$ was $-1$I do not understand what mistake i made in calculating the limit as $x\to -\infty$
Then i guessed i should have made the substitution $x=-t$ and 
$\lim_{x\to-\infty}\sqrt{x^2+ax}-\sqrt{x^2+bx}=\lim_{t\to\infty}\sqrt{t^2-at}-\sqrt{t^2-bt}=\lim_{t\to \infty}\frac{(b-a)t}{\sqrt{t^2-at}+\sqrt{t^2-bt}}=\lim_{t\to\infty}\frac{-2t}{\sqrt{t^2-at}+\sqrt{t^2-bt}}$
$=\lim_{t\to\infty}\frac{-2}{\sqrt{1-\frac{a}{t}}+\sqrt{1-\frac{b}{t}}}=\frac{-2}{2}=-1$

I want to ask why the answer came wrong in the first method and correct in the second method.

Is it always necessary to put $x=-t$ and change the limit to plus
  infinity while calculating the limit as $x\to-\infty$
  Please help me.Thanks.

 A: The short answer to your question is that $x=\sqrt{x^2}$ only for $x > 0$. For $x<0$, we have $x=-\sqrt{x^2}$.
In the first method, the fact that
$$\dfrac{2x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}} = \dfrac2{\sqrt{1+a/x}+\sqrt{1+b/x}}$$
is true only for $x > 0$. For $x<0$, we have $x = -\sqrt{x^2}$, and hence we have
$$\dfrac{2x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}} = -\dfrac2{\sqrt{1+a/x}+\sqrt{1+b/x}}$$
A: Notice, here is an easier method to solve 
let $x=-\large \frac{1}{t}$, $$\lim_{x\to -\infty}(\sqrt{x^2+ax}-\sqrt{x^2+bx})$$
$$=\lim_{t\to 0^+}\left(\sqrt{\left(\frac{-1}{t}\right)^2+a\left(\frac{-1}{t}\right)}-\sqrt{\left(\frac{-1}{t}\right)^2+b\left(\frac{-1}{t}\right)}\right)$$
$$=\lim_{t\to 0^+}\frac{1}{t}\left(\sqrt{1-at}-\sqrt{1-bt}\right)$$
$$=\lim_{t\to 0^+}\frac{\left(\sqrt{1-at}-\sqrt{1-bt}\right)\left(\sqrt{1-at}+\sqrt{1-bt}\right)}{t\left(\sqrt{1-at}+\sqrt{1-bt}\right)}$$
$$=\lim_{t\to 0^+}\frac{1-at-1+bt}{t\left(\sqrt{1-at}+\sqrt{1-bt}\right)}$$
$$=\lim_{t\to 0^+}\frac{b-a}{\sqrt{1-at}+\sqrt{1-bt}}$$
$$=\frac{(b-a)}{1+1}=\color{red}{\frac{b-a}{2}}$$
A: The crucial thing to recall is that $\sqrt{x^2} = |x|$ -- at core, the composition of the square root with the quadratic operator results in something that is basically bi-linear.
Consider $x^2 -3x + 4$; let's first express it as a transformation of $x^2$:
$$x^2-3x+4 = (x-\frac{3}{2})^2 + \frac{7}{4}$$
When we take the square root of this expression, and let $x$ get arbitrarily far away from the vertex of the parabola, the vertical shift will matter less and less. To see this, consider the graphs of $\sqrt{x^2-3x+4}$ and of $\sqrt{x^2-3x+10}$ (i.e., differing only in their constant):

(via Wolfram|Alpha)
As we get further from the underlying vertex ($\frac{3}{2}$), the difference in constants stops mattering -- both look identical to the underlying absolute function, $|x-\frac{3}{2}|$.
Turning to the problem at hand, expressing it like the function above we get:
\begin{align}\sqrt{x^2+ax}-\sqrt{x^2+bx} & =\sqrt{(x+\frac{a}{2})^2-\frac{a^2}{4}}-\sqrt{(x+\frac{b}{2})^2-\frac{b^2}{4}} \\\\ & \approx |x+\frac{a}{2}| - |x+\frac{b}{2}| \\\\& = \frac{b}{2}-\frac{a}{2}\end{align}
The last line follows because when $x$ gets very negative (specifically, for $x < \min \{ -\frac{a}{2}, -\frac{b}{2}\}$), both $x+\frac{a}{2}$ and $x+\frac{b}{2}$ will be negative, so they both flip signs when taking the absolute value.
