Computing the limit with a radical: $ \lim_{x\to\infty}\left(\sqrt{x^2+4x}-x\right) $ I've been struggling with the below problem (I'm a Calc I student and don't know how to approach this using the skills I've developed so far):
$ \lim _{x\to \infty }\left(\sqrt{x^2+4x}-x\right) $
I've tried rewriting using the conjugate:
$ \dfrac{4x}{\sqrt{x^2+4x}+x} $
But I'm not sure how to proceed. Writing the part of the denominator that is under the radical as a power of $ \frac{1}{2} $ is the only thing I can think of, but what emerges is really messy algebraically and I still don't know how to find the limit from there.
 A: Hint: $$ \frac{4x}{\sqrt{x^2+4x}+x} =\frac{4x}{x\sqrt{1+4/x}+x}=\frac{4}{\sqrt{1+4/x}+1}$$
Note, that after taking $x$ out of the root we should have to use $|x|$ but as $x > 0$ it is the same as $x$.
A: $$ \frac{4x}{\sqrt{x^2+4x}+x} =\frac{4x}{|x|\sqrt{1+4/x}+x}$$
Then 
$$\lim_{x \rightarrow +\infty} \frac{4x}{\sqrt{x^2+4x}+x}=\lim_{x \rightarrow +\infty}\frac{4x}{|x|\sqrt{1+4/x}+x}=\frac{4}{1+1}=2$$
$$\lim_{x \rightarrow -\infty} \frac{4x}{\sqrt{x^2+4x}+x}=\lim_{x \rightarrow -\infty}\frac{4x}{|x|\sqrt{1+4/x}+x}=\frac{4}{1-1}=+\infty$$
A: Continuing from where you left
$$ \lim_{x \rightarrow \infty} \frac{4x}{\sqrt{x^2+4x}+x}=\lim_{x \rightarrow \infty} \frac{4}{\sqrt{1+\frac{4}{x}}+1}=2$$
A: When $x$ tends to $+\infty$, simply use equivalents:
$x^2+4x\sim_{+\infty} x^2$, hence $\sqrt{x^2+4x}\sim_{+\infty} \sqrt{x^2}=x$, which means $$\sqrt{x^2+4x}=x+o(x),\quad\text{hence}\quad\sqrt{x^2+4x}+x=2x+o(x),$$so the fraction is 
$$ \frac{4x}{\sqrt{x^2+4x}+x}=\frac{4x}{2x+o(x)}=\frac 2{1+o(1)}\to 2.
$$
When $x$ tends to $-\infty$,  $\sqrt{x^2+4x}\sim_{-\infty} \sqrt{x^2}=\lvert x\rvert$, so (we may suppose $x<0$):
$$\sqrt{x^2+4x}=\lvert x\rvert+o(\lvert x\rvert)-x=2\lvert x\rvert+o(\lvert x\rvert)\sim_{-\infty}2\lvert x\rvert\to+\infty.$$
A: A good trick for solving limits when you get
$$\lim{f(x)}=\frac{'\infty'}{\infty}$$
Or
$$\lim{f(x)}=\frac{'0'}{0}$$
Is to divide the numerator and denominator by the highest power of $x$.  In your case, the highest power is one so you divide top and bottom by $x$ (which is essentially just multiplying by 1).
