Find the limit of the sequence with the general term given. $\displaystyle a_{n}= (\sqrt{4-\frac{1}{n}}-2)n$
I believe it to be 0, but with small confidence.
$\displaystyle \begin{align*}a_n & = \left(\sqrt{4-\dfrac{1}{n}}-2\right)n \\ & = \dfrac{\left(4-\dfrac{1}{n}-4\right)n}{\sqrt{4-\dfrac{1}{n}}+2} \\ & = -\dfrac{1}{\sqrt{4-\dfrac{1}{n}}+2}\end{align*}$
As $\displaystyle n \to \infty$, $\displaystyle \dfrac{1}{n} \to 0$, so:
$\displaystyle \lim_{n \to \infty} a_n = -\dfrac{1}{\sqrt{4}+2} = -\dfrac{1}{4}$