# What will be $\lim_{x \to 0}\sin(1/x)$? [duplicate]

What will be the following limits?

$\lim\limits_{x \to 0}\sin(\frac{1}{x})$

for $x \to 0 ,$ $\frac{1}{x} \to \infty$

Now what will be the $\sin$ value on $\infty$ ?

• The limit $\lim_{x\rightarrow\infty} sin(x)$ does not exists. Therefore, the given limit does not exist either. Jun 6 '16 at 12:08
• Limit doesn't exist I think takeLHL,RHL Jun 6 '16 at 12:09
• In my school days,one teacher explained this in a logical way,rather than mathematical way,but I didn't remember that. Jun 6 '16 at 12:10
• Watch "mean girls" this is covered in there, towards the end (: Jun 6 '16 at 12:11
• Some related posts: math.stackexchange.com/questions/908989/… and math.stackexchange.com/questions/52371/… Jun 6 '16 at 12:51

You say

for $x \to 0 ,$ $\frac{1}{x} \to \infty$

but that's inaccurate. The right-handed limit is indeed $+\infty$, but the left-handed limit will be $-\infty$. But in any case, the limit in question does not exist because both limits $$\lim_{x \to +\infty} \sin x$$ and $$\lim_{x \to -\infty} \sin x$$ do not exist; $\sin x$ will keep oscillating between $-1$ and $1$, so also $$\lim_{x \to 0} \sin \tfrac{1}{x}$$ does not exist.

If you want a formal proof rather than an informal or intuitive reasoning, you can look at a few of the linked answers that cover the same or a similar limit. You'll have to go back to the definition or to useful equivalent statements such as looking at the images of different sequences.

Let $a_n=\dfrac{1}{n\pi}$ and $b_n=\dfrac{1}{\dfrac{\pi}{2}-n\pi}$. We have $$\underset{n\to \infty }{\mathop{\lim }}\,f({{a}_{n}})=0\,\,\,\,\,,\,\,\,\,\underset{n\to \infty }{\mathop{\lim }}\,f({{b}_{n}})=\underset{n\to \infty }{\mathop{\lim }}{{(-1)}^{n}}$$

• @Arnaud D. Thanks. Jun 6 '16 at 13:30

Let me show why the limit does not exist: the reason is the definition of what is a limit (and what is not).

A limit, in first place, is defined as the value where a point $c$ of the domain of the function converges. Then you need to understand what is convergence: it is said that a function $f:\mathcal D\to \Bbb R$ converges to a point $L\in \Bbb R$ when the values of the function get more and more closer to $L$ when you approach to a point $c\in \mathcal D$ with values $x$ more and more closer to $c$.

In more clear words: when you take values $x$ more and more closer to $c$ then the respective $f(x)$ are more and more close to $L$. In mathematical words we said that a function have a limit $L$ at $c$ if for any positive quantity $\varepsilon$ exists a respective positive quantity $\delta$ such that for any $x$ at a distance less than $\delta$ of $c$ then we have that $f(x)$ is at a distance less than $\varepsilon$ of $L$. In mathematical symbols

$$(\forall\varepsilon>0)(\exists\delta>0)(\forall x\in\mathcal D):0<|x-c|<\delta\implies|f(x)-L|<\varepsilon$$

where $\mathcal D$ is the domain of the function. Under these conditions we says that $L=\lim_{x\to c}f(x)$.

For the case of $f(x)=\sin(1/x)$ when we approach to zero taking little and more little values then we get bigger and bigger values inside the sine function and, in any case, the sine function is a periodic function so we never get more closer to a value than other, in other words, the function $\sin(1/x)$ is not converging to any value when we are approaching to $x=0$. Then the limit does not exist, i.e. there are not convergency to some value.

There is no definite value of $\sin$ at infinity. As you might know that infinity is not a number . It is a notion that shows that whatever the value we are considering it is big. So the limit does not exist.

One thing to note here is that the reason of the non-existing limit is not that the number in consideration is going to infinity. The reason is that we don't really know that value of $\sin$ at that angle. The value of sin can be anywhere between $-1$ and $1$.