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Prove that $$\lim_{x \to \infty} \frac{\sin x}{x} =0$$

For a given $\epsilon \gt 0$ we have

\begin{align}\left |\frac{\sin x}{x} -0\right|&\lt \epsilon\\ \implies \frac{|\sin x|}{|x|} &\lt \epsilon\end{align}

From here how can we get an $M \gt 0$ such that $x \gt M$ $\implies$ $|f(x)-L|\lt \epsilon$

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    $\begingroup$ $M=1/ \varepsilon$ $\endgroup$ – Crostul Jul 10 '17 at 9:47
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    $\begingroup$ Clearly $\frac{|\sin x|}{|x|} \leq 1/x$ for $x$ positive and large. Now try to choose $M$ such that $1/x \leq \epsilon$ when $x > M$. $\endgroup$ – Zubzub Jul 10 '17 at 9:52
  • $\begingroup$ ok but how do we know that $\frac{1}{x}$ is less than $\epsilon$, it can also be greater than $\epsilon$ right? $\endgroup$ – Ekaveera Kumar Sharma Jul 10 '17 at 9:59
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Well, the proof goes like this:

Let $\epsilon>0$. We have to find a $M=M(\epsilon)>0$, such that for every $x>M$, we have that $\left|\frac{\sin x}{x}\right|<\epsilon$.

Note that $|\sin x|\leq1$ for every $x\in\mathbb{R}$. Note also that, from Archimedes-Eudoxus Principle, we can find a $n_0=n_0(\epsilon)\in\mathbb{N}$ such that: $$\frac{1}{n_0}<\epsilon$$

Let $M=n_0$. Now, since $\frac{1}{x}$ is strictly decreasing and, hence $x>M\Rightarrow\frac{1}{x}<\frac{1}{M}=\frac{1}{n_0}<\epsilon$, we have, for every $x>M>0$: $$\left|\frac{\sin x}{x}\right|\leq\left|\frac{1}{x}\right|\overset{x>0}{=}\frac{1}{x}<\frac{1}{M}<\epsilon$$ So, the proof is complete.

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The function $\sin x$ always lies between $0$ and $1$. So it will have any of the value between $0-1$, when $x$ is tending to infinity. So as $x$ tends to infinity $\frac{\sin x}{x}$ will be zero.

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  • $\begingroup$ $-1\leq\sin(x)\leq 1$ not $0\leq\sin(x)\leq 1$. $\endgroup$ – Dave Jul 10 '17 at 14:53

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