Question

\[Find \lim_{x \rightarrow \infty} f(x)=\sqrt{4x^2-2x+1}-2x\]

Answer 1

multiply top and bottom by the conjugate of the top

Answer 2

\[\lim_{x \rightarrow \infty}\frac{(4x^2-2x+1)-4x^2}{\sqrt{4x^2-2x+1}+2x}=\lim_{x \rightarrow \infty}\frac{-2x+1}{\sqrt{4x^2-2x+1}+2x}\]
now both top and bottom by sqrt{x^2}
remember sqrt{x^2}=|x|
|x|=x when x>0
|x|=-x when x<0
so since x-> positive infinity we will use sqrt{x^2}=x