# Epsilon N proof of limit of a trigonometric sequence

I'm trying to write a proof by definition (epsilon N) of a limit of a following equation: $$c_n = 2\sin{\frac{1}{2\sqrt{n+1} + \sqrt{n}}}$$ $$\lim_{n \to \infty}c_n = 0$$

Here it is quite intuitive because as n $$\to \infty$$ terms in the denominator tend to $$\infty$$ and limit of sine is $$0$$.

My question is; What do I do with sine function, are we taking invert function $$\arcsin$$ and then modify the $$\varepsilon$$?

$$\left |2\sin{\frac{1}{2\sqrt{n+1} + \sqrt{n}}} - 0\right| <\varepsilon$$

$$\left |2\sin{\frac{1}{2\sqrt{n+1} + \sqrt{n}}}\right| <\varepsilon$$

$$-\varepsilon <2\sin{\frac{1}{2\sqrt{n+1} + \sqrt{n}}} < \varepsilon$$

I'd suggest using inequalities to get rid of the sine: $$\sin(t) for $$t>0$$. Also notice that the argument of the sine: $$\frac{1}{2\sqrt{n+1}+\sqrt{n}}\in(0,1)\Rightarrow \sin\frac{1}{2\sqrt{n+1}+\sqrt{n}}>0$$ so we can also get rid of the absolute value. Hence, \begin{align} \left|2\sin\frac{1}{2\sqrt{n+1}+\sqrt{n}}\right|&=2\sin\frac{1}{2\sqrt{n+1}+\sqrt{n}} \\ &<\frac{2}{2\sqrt{n+1}+\sqrt{n}} \\ &<\frac{2}{2\sqrt{n}+\sqrt{n}}=\frac{2}{3}\frac{1}{\sqrt{n}} \end{align} and we want the last fraction to be less than $$\varepsilon$$. Take $$n>N$$ where $$N=\frac{4}{9\varepsilon^2}$$.
• I just have one more question, how would I go about proving for example that $|3^{n+1} - 5*2^{n-5}|<\epsilon$ Feb 12, 2020 at 21:09