<s>[INST] 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...
 [/INST] 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.
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