Limit of $\sqrt{4x^2 + 3x} - 2x$ as $x \to \infty$

Question

$$\lim_{x\to\infty} \sqrt{4x^2 + 3x} - 2x$$

I thought I could multiply both numerator and denominator by $\frac{1}{x}$, giving

$$\lim_{x\to\infty}\frac{\sqrt{4 + \frac{3}{x}} -2}{\frac{1}{x}}$$

then as x approaches infinity, $\frac{3}{x}$ essentially becomes zero, so we're left with 2-2 in the numerator and $\frac{1}{x}$ in the denominator, which I thought would mean that the limit is zero.

That's apparently wrong and I understand (algebraically) how to solve the problem using the conjugate, but I don't understand what's wrong about the method I tried to use.

Answer 1

Hint: $\sqrt{4x^{2}+3x}-2x=\frac{3x}{\sqrt{4x^{2}+3x}+2x}=\frac{3}{\sqrt{4+\frac{3}{x}}+2}$

Answer 2

Your method yields:

$$\sqrt{4x^2 + 3x} - 2x = \frac{\sqrt{4 + \frac{3}{x}} - 2}{1/x}$$

Hence both the numerator and denominator tend to $0$ as $x \to \infty$.

Answer 3

I always change the variable with $y = \frac{1}{x}$ and take the limit to $y\rightarrow 0$

$$ {\rm limit} = \sqrt{\frac{4}{y^2} + \frac{3}{y}} - \frac{2}{y} = \left. \frac{\sqrt{3 y+4}-2}{y} \right|_{y\rightarrow 0} $$

No with LH rule

$$ {\rm limit} =\left. \frac{3}{2 \sqrt{3 y+4}} \right|_{y\rightarrow 0} = \frac{3}{4} $$

Answer 4

Consider $x\rightarrow\infty$, then $4x^2>>3x$ and define $\epsilon=3x/4x^2<<0$ in that limit. Then $\sqrt{4x^2+3x}-2x=2x\sqrt{1+\epsilon}-2x=2x(1+\epsilon/2)+O(\epsilon^2)-2x=x\epsilon=3/4$. Where you use the expansion for $\sqrt(1+\epsilon)$ around $\epsilon=0$.