How to calculate the limit of $\sqrt{4 x^2 - x} + 2 x$ as $x$ approaches $-\infty$?

Question

$$\lim_{x\to-∞} \sqrt{4 x^2 - x} + 2 x$$ I currently study in 12th grade, and I'm unable to find out how to go about this problem. I've tried multiplying top and bottom with $\sqrt{4 x^2 - x}$, but I'm unable to get at a solution.

Answer 1

You may proceed as follows:

• Set $x = -\frac{1}{h}$ $$\Rightarrow \lim_{x\to-\infty} \sqrt{4 x^2 - x} + 2 x = \lim_{h\to 0^+} \frac{\sqrt{4+h}-2}{h} = \color{blue}{f'(0)} \mbox{ for } \color{blue}{f(h) =\sqrt{4+h}}$$ $$\Rightarrow \boxed{\lim_{x\to-\infty} \sqrt{4 x^2 - x} + 2 x} = \color{blue}{f'(0)} =\frac{1}{2\sqrt{4+0}} \boxed{= \frac{1}{4}}$$

Answer 2

Instead, try multiplying the top and bottom by $\sqrt{4x^2 - x} - 2x$ to find that $$ \lim_{x \to -\infty} (\sqrt{4x^2 - x} + 2x)\cdot \frac{\sqrt{4x^2 - x} - 2x}{\sqrt{4x^2 - x} - 2x} = \\ \lim_{x \to -\infty} \frac{(\sqrt{4x^2 - x})^2 - (2x)^2}{\sqrt{4x^2 - x} - 2x} = \\ \lim_{x \to -\infty} \frac{-x}{\sqrt{4x^2 - x} - 2x} $$ From there, it is helpful to divide the top and bottom by $x$. When rewriting the bottom, it is useful to observe that for $x < 0$ we have $x = -\sqrt{x^2}$.

Answer 3

The other answer gives the standard algebraic trick. But you may also use Taylor series, which gives a lot more quantitative insight. (You may not have covered Taylor series yet, but it is worth seeing another approach.)

Recall from the generalized binomial series that

$$\sqrt{1+u}\color{red}{=}\sum_{k=0}^\infty\binom{1/2}{k}u^k$$

where $\binom{1/2}{k}$ is the generalized binomial coefficient, and the expansion is valid for $|u|<1$.

In your case,

$$\sqrt{4x^2-x}=\sqrt{4x^2\left(1-\frac{1}{4x}\right)}=2|x|\sqrt{1-\frac{1}{4x}}\color{red}{=}2|x|\sum_{k=0}^\infty\binom{1/2}{k}\frac{(-1)^k}{(4x)^k}$$

for $|1/(4x)|<1 \implies |x|>1/4$. Thus the expansion is valid for investigating the limits at infinity.

When $x<0$, we have $|x|=-x$. And since $\binom{1/2}{0}=1$, the expression you are taking the limit of is

$$2x\left(1-\sum_{\color{blue}{k=0}}^\infty\binom{1/2}{k}\frac{(-1)^k}{(4x)^k}\right)=-2x\sum_{\color{blue}{k=1}}^\infty\binom{1/2}{k}\frac{(-1)^k}{(4x)^k}=-2x\left(-\binom{1/2}{1}\frac{1}{4x}+O(x^{-2})\right)$$

which is just $$\frac{1}{2}\binom{1/2}{1}+O(x^{-1})=\frac{1}{4}\,\text{as $x\to-\infty$}$$

Answer 4

Rewrite:

$y:=-x$ , $y >0$ for $x<0.$

$\sqrt{4y^2+y} -2y=$

$2y\sqrt{1+1/4y}-2y=$

$2y(\sqrt{1+1/(4y)} -1)=$

$2y\dfrac{1/(4y)}{\sqrt{1+1/(4y)}+1}=$

$(1/2)\dfrac{1}{\sqrt{1+1/(4y)}+1}.$

Take the limit $y \rightarrow \infty$