# 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 $\frac{5}{4}$?

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.

• Oh, you are totally right, thanks for the help. I really appreciate it. – James Smith Oct 11 '15 at 23:53

\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*}

Expanding DirkGently's answer, this is the point where multiplying by the conjugate is useful:

$\begin{array}\\ \lim_{x\to-\infty}2x\left(1-\sqrt{1-\frac{5}{4x}}\right) &=\lim_{x\to-\infty}2x\left(1-\sqrt{1-\frac{5}{4x}}\right) \frac{1+\sqrt{1-\frac{5}{4x}}}{1+\sqrt{1-\frac{5}{4x}}}\\ &=\lim_{x\to-\infty}2x\frac{\left(1-(1-\frac{5}{4x})\right) }{1+\sqrt{1-\frac{5}{4x}}}\\ &=\lim_{x\to-\infty}2x\frac{\left(\frac{5}{4x})\right) }{1+\sqrt{1-\frac{5}{4x}}}\\ &=\lim_{x\to-\infty}\frac{\frac{5}{2} }{2}\\ &= \frac54\\ \end{array}$

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, $$\lim_{x\to-\infty}\frac{5|x|}{\sqrt{4x^2+5|x|}+2|x|} =\lim_{x\to-\infty}\frac{5}{\sqrt{4+5/|x|}+2}=\frac54$$