$$\lim_{x\to\infty} \sqrt{4x^2 + 3x} – 2x$$
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.
Best Answer
Hint: $\sqrt{4x^{2}+3x}-2x=\frac{3x}{\sqrt{4x^{2}+3x}+2x}=\frac{3}{\sqrt{4+\frac{3}{x}}+2}$