<s>[INST] 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.

 [/INST] 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
$$\dfrac{2x}{\sqrt{x^2+ax}+\sqrt{x^2+bx}} = -\dfrac2{\sqrt{1+a/x}+\sqrt{1+b/x}}$$
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