TITLE: Find an accurate value of $f(x)=\sqrt{4x^2+x}-2x$ for large values of x. Calculate $\lim_{x\to\infty}f(x)$
QUESTION [5 upvotes]: 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...

REPLY [3 votes]: 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 $(1)$