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