find the limit:$\lim_{x\to 0}\frac{\sin\pi/x}{\pi/x}$

Question

Find the limit : $$ \lim_{x\to 0}\frac{\sin\pi/x}{\pi/x}$$

My try:

$$f(x)=\frac{\sin \pi/x}{\pi/x}:D_f=\mathbb{R}-{1/k;k \in\mathbb{Z}}$$

So don't exit limit at to zero

Is it true ?

It will be 0 i suppose if you take x up as sin would just return value btwn $-1,1$ — 2017-02-10
Let $y = \pi / x$, if $x$ goes to zero (doesn't matter from which side), what happens to $y$ and hence to $\sin(y)/y$? — 2017-02-10
The fact that all $1/k$ are excluded from the domain does not imply that there is no limit. There are enough remaining number in between :-) — 2017-02-10
In your related question, http://math.stackexchange.com/questions/2138136/lim-x-to-0-frac-sin-pi-x-sin-pi-x , it was appropriate to exclude the reciprocals of the integers from $D_f$, but there is no need to do so here, since the denominator is nonzero for all nonzero $x$. — 2017-02-10

user id:247327 12k · Accepted Answer

If your "$sin\frac{\pi}{x}$" is intended to be $sin\left(\frac{\pi}{x}\right)$, then let $y= \frac{\pi}{x}$. As x goes to 0, y goes to infinity so this becomes $\lim_{y\to\infty}\frac{sin(y)}{y}$ which is 0 because the numerator stays between -1 and 1 while the denominator goes to infinity.

It is true that the function value does not exist at a countable number of points but that is irrelevant to the limit.

As I point out in a comment below the OP, there is no need here to rule out the reciprocals of the integers. But if for reason you do, it *can be* relevant to the existence of the limit, depending on how you define the concept of limit. — 2017-02-10

user id:104171 17k · Accepted Answer

Take $y=\dfrac{\pi}{x}$ so $$x\to 0 \Rightarrow y\to \infty$$ $$\lim_{x\to 0}\frac{\sin \dfrac{\pi}{x}}{\dfrac{\pi}{x}}=\\ \lim_{y\to \infty}\frac{\sin y}{y} \to 0\\$$

user id:399263 13k · Accepted Answer

$$|\frac{\sin(\frac{\pi}{x})}{\frac{\pi}{x}}|=|x||\frac{\sin(\frac{\pi}{x})}{\pi}|\le|x||\frac{1}{\pi}|\le|x|\to 0$$

user id:384029 12k · Accepted Answer

$$\left|\frac{\sin(\pi/x)}{\pi/x}\right|=\frac{|x|}{\pi}|\sin(\pi/x)|\le \frac{|x|}{\pi}\to 0$$ as $x\to0$ since $|\sin(\pi/x)|\le 1$

user id:202111 6k · Accepted Answer

hint: $$f(x)=\frac{\sin \pi/x}{\pi/x}$$ $$=x/\pi \sin {(\pi/x)}$$ which is always less than or equal to $|x/\pi|$ and always greater than or equal to $-|x/\pi|$.

user id:312594 2k · Accepted Answer

Hint $t=\dfrac{\pi}{x}$

$\frac{-1}{t} \le \frac{sint}{t} \le \frac{1}{t}$

then put $$ t \to \infty$$

find the limit:$\lim_{x\to 0}\frac{\sin\pi/x}{\pi/x}$

6 Answers 6

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