How would you prove that $\lim\limits_{n\to\infty}(\sqrt{4n^2+n}-2n)=\frac14$?

$$\lim_{n\rightarrow \infty}\left[{\sqrt{4n^2+n}-2n}\right]=\frac{1}{4}$$ I am trying to use the definition of the limit but have no idea how to simplify the expression with radical!

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so by the definition, $\forall n > N \rightarrow |a_n - a | < \epsilon, \text{where}\, \ a_n=\sqrt{4n^2+n}-2n\ \, \text{and}\,\ a = \frac{1}{4}$

so after multiply the conjugate and negate $\frac{1}{4}$, I get $\frac{2n-\sqrt{4n^2+n}}{4(\sqrt{4n^2+n}+2n)}$

Since there is still radical in the numerator, I think I have multiply the conjugate again...right?? And then find some formula that is greater!!??? confuse me this real analysis!!

Solution 1:

$$\frac{\sqrt{4n^2+n}-2n}{1}=\frac{\left(\sqrt{4n^2+n}-2n\right)\left(\sqrt{4n^2+n}+2n\right)}{\sqrt{4n^2+n}+2n}=\frac{4n^2+n-4n^2}{\sqrt{4n^2+n}+2n}$$

Can you continue?

Solution 2:

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