TITLE: Help showing $\lim\limits _{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right) = \frac{5}{4}$
QUESTION [5 upvotes]: I am stuck solving the following limit. I know the answer is 5/4, I just can't get it. This is the steps I have done so far.
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right)$
Multiply by Conjugate
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right)\cdot \frac{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Multiply Out
$\lim\limits_{x\to -\infty \:}\cdot \frac{\left(4\cdot \:\:\:x^2-5\cdot \:\:\:x-4\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Combine Like Terms
$\lim\limits_{x\to -\infty \:}\cdot \frac{\left(4\cdot \:\:\:x^2-9\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Factor out x
$\lim\limits_{x\to -\infty \:}\cdot \frac{x\left(4\cdot \:x-9\right)}{\left(\sqrt{x^2\left(4-\frac{5}{x}\right)}-2\cdot \:\:x\right)}$
Pull out x of sqrt and factor again
$\lim \limits_{x\to -\infty \:}\cdot \frac{x\left(4\cdot \:x-9\right)}{x\left(\sqrt{\left(4-\frac{5}{x}\right)}-2\right)}$
Now cancel x terms
$\lim \limits_{x\to -\infty \:}\frac{4\cdot \:\:x-9}{\sqrt{\left(4-\frac{5}{x}\right)}-2}$
Now I don't know what to do next. 
If I plug in I get
$\frac{4\cdot \:\:\:-\infty \:-9}{\sqrt{\left(4-0\right)}-2}=\:\frac{-\infty \:}{0}$ Which doesn't equal $\frac{5}{4}$?

REPLY [2 votes]: Somehow, you missed a square. 
$$
(\sqrt{4x^2-5x})^2-(2x)^2=4x^2-5x-4x^2=-5x
$$
which radically simplifies the ensuing limit calculation, 
$$
\lim_{x\to-\infty}\frac{5|x|}{\sqrt{4x^2+5|x|}+2|x|}
=\lim_{x\to-\infty}\frac{5}{\sqrt{4+5/|x|}+2}=\frac54
$$