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