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I've found this proof for this limit:

$\lim\limits_{x \to + \infty} \frac{sinx}{x}=0$

$|\frac{sinx}{x}-0|=|\frac{sinx}{x}|\le \frac{1}{x}\ltε$ when $x\gt[\frac{1}{ε}]+1$

But I don't seem to understand why we set $x\gt[\frac{1}{ε}]+1$.

Can someone explain this?

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    According to limit definition $\forall\varepsilon>0$, $\exists N\in\mathbb{N}$ such that for $x>N$ we have . . . you choose $N=[\dfrac{1}{ε}]+1$. – 2017-02-12

2 Answers 2

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So, the guess is that $0$ is the limit. Then, according to the definition

$\forall \hspace{0.2cm} \epsilon>0 \hspace{0.2cm}\exists x_0(\epsilon) \in\mathbb{R}|\hspace{0.1cm}|\frac{sinx}{x}-0|<\epsilon\hspace{0.1cm} \forall \hspace{0.1cm} x>x_0(\epsilon)$

So, $x_0(\epsilon)$ is a function of $\epsilon$, which means for any $\epsilon$ that I give you, you should give me $x_0(\epsilon)$, such that $|\frac{sinx}{x}-0|<\epsilon$, for all $x>x_0(\epsilon)$.

According to what you have included

$|\frac{sinx}{x}-0|=|\frac{sinx}{x}|\le \frac{1}{x}$

Then you I give you an $\epsilon$ and if you find an $x_0(\epsilon)$ such that

$\frac{1}{x}<\epsilon \hspace{0.2cm} \forall x>x_0(\epsilon)$

then you automatically have

$|\frac{sinx}{x}-0|<\epsilon \hspace{0.2cm} \forall x>x_0(\epsilon)$

which is required by the definition.

Now, assuming that you know $\epsilon$, you solve the inequality

$\frac{1}{x}<\epsilon \implies x>\frac{1}{\epsilon}$

Now, you can take any $x$ greater than $\frac{1}{\epsilon}$ to be $x_0(\epsilon)$. $x_0(\epsilon)=[\frac{1}{ε}]+1$ is one possibility, which makes $x_0$ to be an integer. But it is not necessary to make it an integer.

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The intention here seems to be that if you choose $\;x>\lfloor\frac1\epsilon\rfloor+1\;$ , then $\;\frac{\sin x}x<\epsilon\;$ , which is what's need to fulfill the definition of limit when $\;x\to\infty\;$ ... (because $\;0<\epsilon\in\Bbb R\;$ is arbitrary and etc.)

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    Yes, but what I don't understand, is how that fulfills this: $\frac{sinx}{x}\ltε$ – 2017-02-12
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    @Michael Because we know (I hope...) that $\;|\sin x|\le 1\;$ ....and thus for positive $\;x\;,\;\;\left|\frac{\sin x}x\right|\le\frac1x\le\epsilon\cdot\;$ etc. – 2017-02-12