Find the oblique asymptote of $\sqrt{x^2+3x}$ for $x\rightarrow-\infty$ I want to find the asymptote oblique of the following function for $x\rightarrow\pm\infty$
$$f(x)=\sqrt{x^2+3x}=\sqrt{x^2\left(1+\frac{3x}{x^2}\right)}\sim\sqrt{x^2}=|x|$$
For $x\rightarrow+\infty$ we have:
$$\frac{f(x)}{x}\sim\frac{|x|}{x}=\frac{x}{x}=1$$
which means that the function grows linearly. 
$$f(x)-mx=\sqrt{x^2+3x}-x=x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}-1\right)\sim x\left(\frac {1}{2}\cdot\frac{3}{x}\right)=\frac{3}{2}$$
The oblique asymptote is $y=x+\frac 3 2$ which is correct. For $x\rightarrow-\infty$ we have:
$$\frac{f(x)}{x}=\frac{|x|}{x}=\frac{-x}{x}=-1$$
This means that 
$$f(x)-mx=\sqrt{x^2+3x}+x=x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}+1\right)=x\left(\sqrt{1+\frac{3}{x}}+1\right)\sim x\cdot2\rightarrow -\infty$$
Which is not what my textbook reports ($-\frac{3}{2}$). Any hints on what I did wrong to find the $q$ for $x\rightarrow-\infty$?

The problem is in
$$
f(x)-mx=\sqrt{x^2+3x}+x\underset{\begin{array}{c} \uparrow \\ \text{problem}\end{array}}{=}x\left(\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}+1\right)=x\left(\sqrt{1+\frac{3}{x}}+1\right)\sim x\cdot2\rightarrow -\infty
$$
which should be
$$
f(x)-mx=\sqrt{x^2+3x}+x=x\left(-\sqrt{\frac{x^2}{x^2}+\frac{3x}{x^2}}+1\right)
$$
because we are near $-\infty$, so $\sqrt{x^2}=-x$.

In order to avoid such common mistakes, I suggest to change $x=-t$, when the limit for $x\to-\infty$ is involved:
$$
\lim_{x\to-\infty}\bigl(\sqrt{x^2+3x}+x\bigr)=
\lim_{t\to\infty}\bigl(\sqrt{t^2-3t}-t\bigr)=
\lim_{t\to\infty}\frac{-3t}{\sqrt{t^2-3t}+t}
$$
From here it should be clear.
Another strategy, in this case, is to set $x=-1/t$, that transforms the limit into
$$
\lim_{t\to0^+}\left(\sqrt{\frac{1}{t^2}-\frac{3}{t}}-\frac{1}{t}\right)=
\lim_{t\to0^+}\frac{\sqrt{1-3t}-1}{t}
$$
which is a simple derivative: if $f(t)=\sqrt{1-3t}$, then $f'(t)=\frac{-3}{2\sqrt{1-3t}}$ and
$$
\lim_{t\to0^+}\frac{\sqrt{1-3t}-1}{t}=f'(0)=-\frac{3}{2}
$$
