# Epsilon-Delta Proof $\lim\limits_{n\to\infty}\frac{n+\sin(n)}{n+1} = 1$ [duplicate]

How To Construct a Epsilon-Delta proof $\displaystyle \lim\limits_{n\to\infty}\frac{n+\sin(n)}{n+1} = 1$ ?

The beginning:

Fix $\epsilon$ > 0. Is there an $N\in \mathbb{N},$ such that $$n\ge N \implies\left|\frac{n+\sin(n)}{n+1}-1 \right|<\epsilon?$$

$$\left|\frac{n+\sin(n)}{n+1}-1 \right|<\epsilon \iff$$ $$\left|\frac{\sin(n)-1}{n+1}\right|<\epsilon \iff$$ $$\ldots$$

I got that $n > \frac{2}{\epsilon}-1$, is that right?

N=(max or min ?) {0, $\lfloor\frac{2}{\epsilon}-1\rfloor$}

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• So far so good. Note that the numerator is bounded. – Matthew Leingang Sep 24 '17 at 10:20
• $|\sin n-1|\leq2$. – Nosrati Sep 24 '17 at 10:21
• No \displaystyle in titles, please. – Did Sep 24 '17 at 15:01

Hint: Use the triangular inequality for the numerator and the boundedness of $\sin(n)$ by $|\sin(n)|\leq 1$:
$$\biggl|\frac{\sin(n)-1}{n+1}\biggr|\leq \frac{|\sin(n)|+1}{|n+1|}\leq \frac{1+1}{|n+1|}=\frac{2}{n+1}$$
• Minor addendum but we could even skip the triangle equality step and say $\left|\frac{\sin(n)-1}{\ldots}\right|=\frac{|\sin(n)-1|}{|\ldots|}\leq\frac{2}{|\ldots|}$ since we know that $-2\leq\sin(n)-1\leq0$. Great answer regardless :) – Jam Sep 24 '17 at 10:37
• @Jam: You are right, but I would still prefer the triangular inequality because it makes it explicit what is implicitly hidden in $-2\leq \sin(n) -1 \leq 0$. – MrYouMath Sep 24 '17 at 11:11
• You could also use $1- \sin(n) = 2\sin^2(n/2 + \pi/4)$ to derive that inequality. – eyeballfrog Sep 24 '17 at 14:29
Try these: $|x-y| \leq |x| + |y|$ and $|\sin(x)|\leq1$.