Evaluate $\lim_{x\to -\infty} (4x^2-x)^{1/2}+2x$

Question

Evaluate $$\lim_{x\to -\infty} (4x^2-x)^{1/2}+2x$$

My try:- $\lim_{x\to -\infty} (4x^2-x)^{1/2}+2x$ = $\lim_{x\to -\infty} (4x^2-x-4x^2)/ ((4x^2-x)^{1/2}-2x)$=-$\infty$

I know this is incorrect as I have seen its graph which tends to 0.25. This means I have done a mistake. But I am unable to find it out.

Answer 1

$$\lim_{x\to-\infty}\sqrt{4x^2-x}+2x=\lim_{x\to-\infty}\frac{x}{2x-\sqrt{4x^2-x}}=\lim_{x\to-\infty}\frac{1}{2+\sqrt{4-1/x}}=\frac{1}{4}$$ $\textbf{Addendum:}$ The last step in the limit follows from the following $$2-\sqrt{4x^2-x}/x=2+\sqrt{4x^2-x}/|x|=2+\sqrt{4x^2/x^2-x/x^2}=2+\sqrt{4-1/x}$$ whenever $x<0$.

Answer 2

Note that for $x<0$ $$(4x^2-x)^{1/2}=-2x\left(1-\frac1{4x}\right)^{1/2}=-2x\left(1-\frac1{8x}+O(|x|^{-2})\right) $$ etc.

Answer 3

Let $-\dfrac1x=h\implies h\to0^+, h>0$

$$4x^2-x=\dfrac{4-h}{h^2}\implies\sqrt{4x^2-x}=\dfrac{\sqrt{4-h}}h$$

as $\sqrt{h^2}=|h|=+h$ as $h>0$

$$\lim_{x\to-\infty}(\sqrt{4x^2-x}+2x)=\lim_{h\to0^+}\dfrac{\sqrt{4-h}-2}h=\lim_{h\to0^+}\dfrac{4-h-4}h\cdot\dfrac1{\lim_{h\to0^+}(\sqrt{4-h}+2)}=?$$

Answer 4

Let $y=-x\to \infty$

$$\lim_{x\to -\infty} \sqrt{4x^2-x}+2x=\lim_{y\to \infty} \sqrt{4y^2+y}-2y$$

and

$$\sqrt{4y^2+y}-2y=\left(\sqrt{4y^2+y}-2y\right)\frac{\sqrt{4y^2+y}+2y}{\sqrt{4y^2+y}+2y}=\frac{4y^2+y-4y^2}{\sqrt{4y^2+y}+2y}\to \frac14$$

Answer 5

$(4x^2-x-4x^2)/(4x^2-x)^{1/2}=-(4x^2-x)^{1/2}/(4x-1)$, not $(4x^2-x)^{1/2}$.

Hint. Substitute $t=-x$, and multiply $\frac{(4t^2+t)^{1/2}+2t}{(4t^2+t)^{1/2}+2t}$.

Answer 6

Your method will also work: $$\begin{align} &\lim_\limits{x\to -\infty} \frac{-x}{\sqrt{4x^2-x}-2x}=\\ &\lim_\limits{x\to -\infty} \frac{-x}{-x\sqrt{4-\frac{1}{x}}-2x}=\\ &\lim_\limits{x\to -\infty} \frac{1}{\sqrt{4-\frac{1}{x}}+2}=\\ &\frac14.\end{align}$$