<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] Notice, here is an easier method to solve 
let $x=-\large \frac{1}{t}$, $$\lim_{x\to -\infty}(\sqrt{x^2+ax}-\sqrt{x^2+bx})$$
$$=\lim_{t\to 0^+}\left(\sqrt{\left(\frac{-1}{t}\right)^2+a\left(\frac{-1}{t}\right)}-\sqrt{\left(\frac{-1}{t}\right)^2+b\left(\frac{-1}{t}\right)}\right)$$
$$=\lim_{t\to 0^+}\frac{1}{t}\left(\sqrt{1-at}-\sqrt{1-bt}\right)$$
$$=\lim_{t\to 0^+}\frac{\left(\sqrt{1-at}-\sqrt{1-bt}\right)\left(\sqrt{1-at}+\sqrt{1-bt}\right)}{t\left(\sqrt{1-at}+\sqrt{1-bt}\right)}$$
$$=\lim_{t\to 0^+}\frac{1-at-1+bt}{t\left(\sqrt{1-at}+\sqrt{1-bt}\right)}$$
$$=\lim_{t\to 0^+}\frac{b-a}{\sqrt{1-at}+\sqrt{1-bt}}$$
$$=\frac{(b-a)}{1+1}=\color{red}{\frac{b-a}{2}}$$
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