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 $x \gt M$ $\implies$ $|f(x)-L|\lt \epsilon$

  • 3
    $\begingroup$ $M=1/ \varepsilon$ $\endgroup$ – Crostul Jul 10 '17 at 9:47
  • 1
    $\begingroup$ 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

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.


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.

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

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.