Value of $\lim\limits_{x\to-\infty}{(4x^2-x)^{1/2} +2x}$ The value of the $$\lim\limits_{x\to-\infty}{(4x^2-x)^{1/2} +2x}$$ is?
The answer given is $1/4$.
I rationalized and got $$\lim\limits_{x\to-\infty} \frac {-x}{|x|[(4- \frac {1}{x})^{1/2}-2]}$$ how to proceed further?
 A: A more elementary solution :
$$\lim_{x\to -\infty} \big( \sqrt{4x^2-x} +2x \big) =\lim_{x\to -\infty} \frac{-x}{\sqrt{4x^2-x} -2x} =\lim_{x\to \infty} \frac{x}{\sqrt{4x^2+x}+2x} =\lim_{x\to\infty} \frac{1}{\sqrt{4+\frac{1}{x}}+2} = \frac{1}{\sqrt{4}+2}=\frac{1}{4}$$
where I used that
$$\lim_{x\to -\infty} f(x)=\lim_{x\to \infty} f(-x)$$
A: The limit is equivalent to
$$\begin{align}
\lim_{x\to-\infty}\left(2|x|\left(1-\frac1{4x}\right)^{1/2}+2x\right)
&=\lim_{x\to-\infty}\left(-2x\left(1-\frac1{4x}\right)^{1/2}+2x\right)\\
&=\lim_{x\to-\infty}-2x\left(\left(1-\frac1{4x}\right)^{1/2}-1\right)\\
\end{align}$$
Then using the generalized binomial expansion we get that as $x\to0$
$$(1+x)^n=1+nx+o(x)$$
Hence our limit becomes
$$\begin{align}
\lim_{x\to-\infty}-2x\left(\left(1-\frac1{4x}\right)^{1/2}-1\right)
&=\lim_{x\to-\infty}-2x\left(1-\frac1{8x}+o\left(\frac1x\right)-1\right)\\
&=\lim_{x\to-\infty}-2x\left(-\frac1{8x}+o\left(\frac1x\right)\right)\\
&=\lim_{x\to-\infty}\left(\frac14+o(1)\right)\\
&=\frac14\\
\end{align}$$
A: \begin{align*}
(4x^2-x)^{1/2}+2x
&=\sqrt{4x^2-x}+2x
\\&=\frac{(\sqrt{4x^2-x}+2x)(\sqrt{4x^2-x}-2x)}{\sqrt{4x^2-x}-2x}
\\&=\frac{(4x^2-x)-4x^2}{\sqrt{4x^2-x}-2x}
\\&=\frac{4x^2-x-4x^2}{\sqrt{4x^2-x}-2x}
\\&=\frac{-x}{\sqrt{4x^2-x}-2x}
\\&=\frac{-x}{\sqrt{(2x)^2(1-\frac{1}{4x})}-2x}
\\&=\frac{-x}{2|x|\sqrt{1-\frac{1}{4x}}-2x}
\\&\qquad [x<0]
\\&=\frac{-x}{-2x\sqrt{1-\frac{1}{4x}}-2x}
\\&=\frac{1}{2\sqrt{1-\frac{1}{4x}}+2}
\\&\to\frac{1}{2\sqrt{1+0}+2}
\\&=\frac{1}{4}.
\end{align*}
(as $x\to-\infty$)
A: Your rationalization is almost correct. It should be
$$\frac{-x}{\sqrt{4x^2-x}-2x} = \frac{-x}{|x|\left[\left(4-\frac{1}{x}\right)^{1/2}-\frac{2x}{|x|}\right]}.$$
You have assumed that $-2x = -2|x|,$ which is not true when $x$ is negative. But as $x\to-\infty,$ we can asume $x<0$ and thus $|x|=-x,$ so this is equal to
$$\frac{1}{\left(4-\frac1x\right)^{1/2}+2}.$$
A: Hint: Multiply by ${\sqrt{4x^2-x}-2x}\over {\sqrt{4x^2-x}-2x}$
The result is ${-x\over{\sqrt{4x^2-x}-2x}}$
$={{-x}\over {|x|\sqrt{4-{1\over x}}+2}}=$
${{-x}\over {-x\sqrt{4-{1\over x}}+2}}$
${1\over {\sqrt{4-{1\over x}}+2}}$.
A: $y:=-x$, and $ \lim y \rightarrow +\infty$.
$(4y^2+y)^{1/2}-2y=$
$((2y+1/4)^2-1/16)^{1/2}-2y$;
$z:=2y+1/4$;
We get
$((z^2-1/16)^{1/2}-z) +1/4$;
Since 
$\lim_{z \rightarrow \infty} (z^2-1/16)^{1/2}-z)=0$ (why?), and we are done.
Note: 
$(z^2-1/16)^{1/2}-(z^2)^{1/2}= $
$\dfrac{-1/16}{(z^2-1/16)^{1/2}+(z^2)^{1/2}}$
A: For the sake of comfort, we change the sign and evaluate
$$\lim\limits_{x\to\infty}{(4x^2+x)^{1/2}-2x}$$
which is
$$\lim\limits_{x\to\infty}\frac{x}{{(4x^2+x)^{1/2}+2x}}$$
or
$$\lim\limits_{x\to\infty}\frac{1}{{(4+\frac1x)^{1/2}+2}}=\frac14.$$
