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Evaluate the following limit: $$\displaystyle\lim_{n\to\infty}(\sqrt{4n^2+n}-2n)$$ So far I've come up with this: $\displaystyle\lim_{n\to\infty}(\sqrt{4n^2+n}-2n)$ = $\displaystyle\lim_{n\to\infty}(\sqrt{4n^2(1+\frac{1}{4n}})-2n)$ = $\displaystyle\lim_{n\to\infty}(2n\sqrt{(1+\frac{1}{4n}})-2n)$ = $\displaystyle\lim_{n\to\infty}(2n(\sqrt{(1+\frac{1}{4n}})-1))$. I think it's pretty clear from here that this goes to infinity, but how can I justify that the 2n grows stronger to infinity than the part in the brackets goes to zero? I know standard rules about exponential functions growing harder than polynomials, but not about this.

Martin Sleziak
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1 Answers1

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The reasoning is not correct, and the limit is not infinity. Instead, multiply by the conjugate on top and bottom:

\begin{align*} \sqrt{4n^2 + n } - 2n &= \Big(\sqrt{4n^2 + n} - 2n\Big) \frac{\sqrt{4n^2 + n} + 2n}{\sqrt{4n^2 + n} + 2n} \\ &= \frac{n}{\sqrt{4n^2 + n} + 2n} \\ &= \frac{1}{\sqrt{4 + \frac 1 n}+2} \end{align*}