Limit with lower-bound

Question

$$\lim _{n→∞}{\frac{\left\lfloor \sqrt{2n - 2}\right\rfloor}{2\sqrt{n}+\cos{n}}}$$ How can I calculate this limit? I can calculate it without the lower bound $\left(\frac{1}2\right)$ but do not know how to do it with it.

Answer 1

Consider the following: \begin{equation} \begin{split} \frac{\sqrt{2n-2}-1}{2\sqrt{n}-\cos(n)}\leq\frac{{\lfloor\sqrt{2n-2}\rfloor}}{2\sqrt{n}-\cos(n)}\leq \frac{\sqrt{2n-2}}{2\sqrt{n}-\cos(n)}\\ \Rightarrow \lim_{n\rightarrow \infty}\frac{\sqrt{2n-2}-1}{2\sqrt{n}-\cos(n)}\leq\lim_{n\rightarrow \infty}\frac{{\lfloor\sqrt{2n-2}\rfloor}}{2\sqrt{n}-\cos(n)}\leq \lim_{n\rightarrow \infty}\frac{\sqrt{2n-2}}{2\sqrt{n}-\cos(n)}\\ \Rightarrow \frac{\sqrt{2}}{2}\leq\lim_{n\rightarrow \infty}\frac{{\lfloor\sqrt{2n-2}\rfloor}}{2\sqrt{n}-\cos(n)}\leq \frac{\sqrt{2}}{2} \end{split} \end{equation} Via the Squeeze Theorem.