Why is $\lim\limits_{x\to0}x\sin(1/x)=0$?

How $\lim\limits_{x\to0} x\sin(1/x)=0$ When $\lim\limits_{x\to0}\sin(1/x)$ doesn't exist? I got this result by calculating it manualy and by wolphram alpha too. I took $-1\le\sin(1/x)\le1$ and by multiplying it by $x$ i got $0$. But still $\lim\limits_{x\to0}\sin(1/x)$ doesn't exist.

Solution 1:

I believe this is the source of your confusion. When you look at $\lim_{x\rightarrow 0}\ x\ \text{sin}(1/x)$, which does exist, you try to apply the limit product rule, which would give $$ \lim_{x\rightarrow 0}\ x\ \text{sin}(1/x) = \left(\lim_{x\rightarrow 0}\ x\right)\left( \lim_{x\rightarrow 0}\ \text{sin}(1/x)\right)$$

The above statement is false because $\lim_{x\rightarrow 0}\ \text{sin}(1/x)$ does not exist. Now, where did we go wrong to arrive at a false statement? We applied the product rule when it could not be applied. Theorem conditions are VERY important. The product rule says that IF the separate limits $\lim_{x\rightarrow c}\ f(x)$ and $\lim_{x\rightarrow c}\ g(x)$ exist, THEN you can apply the product rule $\lim_{x\rightarrow c}\ f(x)g(x)=\left(\lim_{x\rightarrow c}\ f(x)\right)(\lim_{x\rightarrow c}\ g(x))$. The condition is not met in your example, so it is perfectly ok for the product rule result not to hold.