# Can you find $\lim_{x\to0}x^2\sin\left(\frac{1}{x}\right)$ without sandwich theorem?

The lesson I'm doing is solving $$\lim_{x\to0}x^2\sin\left(\dfrac{1}{x}\right)$$ with sandwich theorem like so : $$\displaystyle\lim_{x\to0}-x^2\leq\displaystyle\lim_{x\to0}x^2\sin\left(\dfrac{1}{x}\right)\leq\displaystyle\lim_{x\to 0}x^2$$ When I first look at this though my thought was to use the properties of lim and do this :
$$\lim_{x\to0}x^2 = 0$$ and times that by $$\lim_{x\to0}\sin\left(\dfrac{1}{x}\right)$$
And if I'm not an idiot then this says that $$0\sin(1/0)$$ is going to equal $$0$$ which is the right answer.
Does that make sense? Instead of bothering with sandwich theorem you could just pull apart the $$x^2$$ and $$\sin(1/x)$$, input the $$0$$ of the limit and not even have to bother with the sine.

• the trouble is $\lim_{x\to 0}\sin(1/x)$ is undefined. Sep 8, 2019 at 23:31
• Right but can't you "skip" that since it will be multiplied by 0? That would seem logical to me. Besides in the lessons they are saying dividing by 0 gives infinity. Sep 8, 2019 at 23:32
• Note that properties of limits don't apply here as one of the factors $x^2$ tends to $0$ and the limit of other factor $\sin(1/x)$ does not exist. So you can either go for sandwich theorem or just use the definition of limit directly. Sep 9, 2019 at 9:48

No, this doesn't work. As

$$\lim_{x\to0}\sin\left(\frac1x\right)$$ does not exist, you may not invoke the product rule.

Using the same reasoning, we could write

$$\lim_{x\to0}\frac xx=\left(\lim_{x\to0}x\right)\left(\lim_{x\to0}\frac 1x\right)\color{red}{=0}$$ "because the first limit is $$0$$ and we don't care that the second does not exist".

If you claim that "but it will work because the sine is bounded", you are in fact implicitly using a sandwich argument:

$$a<\sin x then

$$ax^2 proves the limit $$0$$.

The problem is that $$\lim\limits_{x\rightarrow 0}\sin(1/x)=\lim\limits_{X\rightarrow +\infty}\sin(X)$$ doesn't exist. You can also write that $$|x^2\sin(1/x)|\leqslant x^2\underset{x\rightarrow 0}{\longrightarrow} 0$$

• But $0 * sin(infinity)$ would equal zero. Doesn't matter if sin(infinity) doesn't really exist, it must be between -1 and 1 right? Sep 8, 2019 at 23:43
• The fact that $\sin$ is between $-1$ and $1$ is used in the sandwich theorem, what you want to do, the sandwich theorem does it rigorously. You can't manipulate things that doesn't exist... For instance $\lim\limits_{x\rightarrow +\infty}{(\cos x-\cos x)}=0$ but you can't write that $\lim\limits_{x\rightarrow +\infty}{(\cos x-\cos x)}=\lim\limits_{x\rightarrow +\infty}{\cos x}-\lim\limits_{x\rightarrow +\infty}{\cos x}$ because it doesn't make sense, as $1+1=3$ doesn't make sense either. Sep 8, 2019 at 23:48
• What do you mean? That would make perfect sense. infinity = infinity and cos(x) - cos(x) = 0. And besides 0 x anything is 0. so 0 x sin(any number) = 0 Sep 8, 2019 at 23:51
• Infinity is unpredictable, in fact it is not a number so the statement $0\times\sin(any\ number)=0$ does not apply here. Sep 8, 2019 at 23:54
• @DerekC. $0\cdot\text{anything}=0$ is wrong.
– user65203
Sep 10, 2019 at 9:25

I think It make sense. see whatever is in sine function it will always give a value from [-1,1] and if we multiply it by a number which is almost zero we will get zero.

• This is a restatement of the sandwich theorem in disguise.
– user65203
Sep 10, 2019 at 8:49