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 $\varepsilon$?

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

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

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


1 Answer 1


I'd suggest using inequalities to get rid of the sine: $\sin(t)<t$ 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 $N=\frac{4}{9\varepsilon^2}$.

  • $\begingroup$ I just have one more question, how would I go about proving for example that $|3^{n+1} - 5*2^{n-5}|<\epsilon$ $\endgroup$
    – user77723
    Feb 12, 2020 at 21:09
  • $\begingroup$ @user77723 I don't see how this is related to the problem or my answer. If you have a new question, you should post it separately. $\endgroup$
    – bjorn93
    Feb 12, 2020 at 21:10

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