Find an accurate value of $f(x)=\sqrt{4x^2+x}-2x$ for large values of x. Calculate $\lim_{x\to\infty}f(x)$ My works:
$x^2$ can be very large if x is large, thus the function has lose-of-significance error and we need to reformulate it. $$ f(x)=\sqrt{4x^2+x}-2x=\sqrt{x(4x+1)}-2\sqrt{x}\sqrt{x}=\sqrt{x}(\sqrt{4x+1}-2\sqrt{x})$$ The latter will not have any lose-of-significance error in the evaluation.
I am not sure if I did this correctly and I do not know how to compute the limit...
 A: I would factor out the $2x$ from both terms, and at a crucial point use the Binomial expansion for $(1+a)^{1/2}$:
\begin{align}
\sqrt{4x^2+x}-2x&=2x\left[\frac{\sqrt{4x^2+x}}{\sqrt{4x^2}}-1\right]\\
&=2x\left[\sqrt{1+\frac1{4x}}-1\right]\\
&=2x\left[-1+1+\frac12\cdot\frac1{4x}-\frac18\left(\frac{1}{4x}\right)^2+\cdots\right]\\
&=2x\left[ \frac1{8x}-\frac1{128x^2}+\cdots \right]\\
&=1/4-\text{(small)}
\end{align}
A: Setting $\frac1x=h,$
$$\lim_{x\to\infty^+}\sqrt{4x^2+x}-2x=\lim_{h\to0}\frac{\sqrt{4+h}-2}h\ \ \ \ (1)$$
Method $1:$
Now, $$\frac{\sqrt{4+h}-2}h=\frac{(4+h)-2^2}{h(\sqrt{4+h}+2)}=\frac1{\sqrt{4+h}+2}\text{ if }h\ne0$$
Here  as $h\to0,h\ne0$
Can you take it from here?
Method $2:$
This has strong resemblance with your method 
$\displaystyle \sqrt{4+h}=2\left(1+\frac h4\right)^{\frac12}=2\left(1+\frac12\cdot\frac h4+O(h^2)\right)$ (Using Maclaurin series )
$\displaystyle\implies \sqrt{4+h}-2=\frac h4+O(h^2) $
Now, I leave the rest for you to complete
Method $3:$
$$\lim_{h\to0}\frac{\sqrt{4+h}-2}h=\lim_{h\to0}\frac{\sqrt{4+h}-\sqrt4}h=\frac{d(\sqrt x)}{dx}_{(\text{ at } x=4)}=\frac1{2\sqrt x}_{(\text{ at } x=4)}$$
Method $4:$
Apply L'Hôpital's rule on $(1)$
A: Note that
$$\sqrt{4x^2+x}-2x = \dfrac{\sqrt{4x^2+x}+2x}{\sqrt{4x^2+x}+2x} \cdot \sqrt{4x^2+x}-2x = \dfrac{4x^2+x-4x^2}{\sqrt{4x^2+x}+2x} = \dfrac{x}{\sqrt{4x^2+x}+2x}$$
Hence, use $\dfrac{x}{\sqrt{4x^2+x}+2x}$ for computation purposes, especially when $x$ is large. There is no catastrophic cancellation of digits as opposed to evaluating $\sqrt{4x^2+x}-2x$ directly.
