Help showing $\lim\limits _{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right) = \frac{5}{4}$ 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 $\frac{5}{4}$?
 A: \begin{align*}
\lim_{x\to-\infty}\sqrt{4x^2-5x}+2x&=\lim_{x\to-\infty}-2x\sqrt{1-\frac{5}{4x}}+2x\\
&=\lim_{x\to-\infty}2x\left(1-\sqrt{1-\frac{5}{4x}}\right)\\
&=\lim_{x\to-\infty}2x\cdot \frac{\frac5{4x}}{1+\sqrt{1-\frac{5}{4x}}}\\
&=5/4.
\end{align*}
A: Hint: You can use the fact that
$$
\sqrt{4-5t} = 2\sqrt{1-\frac{5}{4}t} = 2\left(1-\frac{5}{8}t + o(t)\right)
$$
when $t\to 0$. (This is the Taylor approximation of $\sqrt{1+t}$ around $0$.) Note that when $x\to-\infty$, $t=\frac{1}{x}\to 0$. 
Additionally, you have a few issues in your derivation. For instance, $x\to -\infty$, so in particular is negative:
$$
\sqrt{x^2} = \lvert x\rvert = -x
$$
(when you factor). Also, even before, when you multiply by the conjugate you should have obtained a $-4x^2$, not $-4x$, in the numerator.
A: Expanding DirkGently's answer,
this is the point
where multiplying
by the conjugate
is useful:
$\begin{array}\\
\lim_{x\to-\infty}2x\left(1-\sqrt{1-\frac{5}{4x}}\right)
&=\lim_{x\to-\infty}2x\left(1-\sqrt{1-\frac{5}{4x}}\right)
\frac{1+\sqrt{1-\frac{5}{4x}}}{1+\sqrt{1-\frac{5}{4x}}}\\
&=\lim_{x\to-\infty}2x\frac{\left(1-(1-\frac{5}{4x})\right)
}{1+\sqrt{1-\frac{5}{4x}}}\\
&=\lim_{x\to-\infty}2x\frac{\left(\frac{5}{4x})\right)
}{1+\sqrt{1-\frac{5}{4x}}}\\
&=\lim_{x\to-\infty}\frac{\frac{5}{2}
}{2}\\
&= \frac54\\
\end{array}
$
A: 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, 
$$
\lim_{x\to-\infty}\frac{5|x|}{\sqrt{4x^2+5|x|}+2|x|}
=\lim_{x\to-\infty}\frac{5}{\sqrt{4+5/|x|}+2}=\frac54
$$
