$\lim_{x\to \infty} [{\sqrt{(4x^2-2x)}+2x}]=[1/4]$ why?

Question

I have this sum: $$\lim_{x\to \infty} \left(\sqrt{4x^2-x}+2x\right)=\dfrac 14$$ in a competitive examination. But they give the answer as $\dfrac 14$. As I have learnt, all I have to worry about limits that how I can eliminate the zero in the denominator. But here I haven't had a perfect limit and since $x\to\infty$ the denominator is zero but how can I get the numerator. Would you please tell me how to start the problem ?

Answer 1

Note that

$$\sqrt{4x^2-2x}+2x\sim 4x \to +\infty$$

while for $\sqrt{4x^2-2x}-2x$ note that by binomial expansion

• $\sqrt{4x^2-2x}=2x\left(1-\frac1{2x}\right)^{\frac12}\sim 2x\left(1-\frac1{4x}\right)=2x-\frac12$

and thus

$$\sqrt{4x^2-2x}-2x\sim 2x-\frac12-2x\to -\frac12$$