0
$\begingroup$

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 $\varepsilon$?

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

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

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

$\endgroup$

1 Answer 1

0
$\begingroup$

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 $N=\frac{4}{9\varepsilon^2}$.

$\endgroup$
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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.