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

Question

$$\lim_{n\rightarrow \infty}\left[{\sqrt{4n^2+n}-2n}\right]=\frac{1}{4}$$ 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!!

Answer 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?

Answer 2

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

Answer 3

Here is a proof I like but is not too easy to generalize:

Let $f(x) = \sqrt{4+x}$. Note that $f(0) = 2$. Now $\lim_{x \rightarrow 0}(\frac{f(x) -f(0)}{x}) = f'(0) = \frac{1}{2\sqrt{4+0}} = \frac{1}{4}$. But also $\lim_{x \rightarrow 0}(\frac{f(x) -f(0)}{x}) = \lim_{x \rightarrow 0}(\frac{\sqrt{4+x}-2}{x})$ by the definition of $f(x)$, and this is also equal to $\lim_{x \rightarrow 0}(\sqrt{\frac{4}{x^2} + \frac{1}{x}} - \frac{2}{x})$. Now when we let $n = \frac{1}{x}$ we recover the original limit $\lim_{n \rightarrow \infty}(\sqrt{4n^2 + n}-2n)$. Hooray!

Answer 4

I would prove it with Taylor expansion:

$\lim_{n\to\infty} \sqrt{4n^2+n}-2n = \lim_{n\to\infty} 2n(-1+\sqrt{1+\frac{4}{n}})=\lim_{n\to\infty}2n(-1 + 1 +\frac{1}{2}\frac{1}{4n}-\frac{1}{8}(\frac{1}{4n})^2+\dots) = \frac{1}{4}$