I have a query about $\lim_{x\to 0} x\sin\left(\frac{1}{x}\right)$.
I know this limit equals 0.
But if I use $\lim f(x)g(x)=\lim f(x)\lim g(x)$,and $\lim_{x\to 0}\sin \frac{1}{x}=\infty$ then I have,
$$\begin{aligned}\lim_{x\to 0} x\sin\left(\frac{1}{x}\right)&=\lim_{x\to 0}x\cdot\lim_{x\to 0}\sin \left(\frac{1}{x}\right)\\&=0\cdot \infty\\&=\infty \end{aligned}$$
Please tell me what is the issue with my this calculation of the limit.
Two things to note here:
First, $\lim_{x\rightarrow 0} \sin\left(\frac{1}{x}\right)$ does not exist, which is evident if you plot it out.
Second, the formula $\lim_{x\rightarrow a} f(x)g(x)=\lim_{x\rightarrow a} f(x) \lim_{x\rightarrow a} g(x)$ works under the assumptions that $\lim_{x\rightarrow a} f(x)$ and $\lim_{x\rightarrow a} g(x)$ both exist (whether infinite or finite), and that you won't have the $\infty\cdot0$ or $0\cdot\infty$ situation.
In particular, since $\lim_{x\rightarrow 0} \sin\left(\frac{1}{x}\right)$ does not exist, you cannot use this formula and you have to find another way to compute your original limit.
According to the Product Law, if $\lim_{x \to a} f(x) = y_1$ and $\lim_{x \to a} g(x) = y_2$ then $\lim_{x \to a} f(x)g(x) = y_1y_2$. If the conditions are met we can be sure that the conclusion is true. Otherwise, this theorem is silent about the $\lim_{x \to a} f(x)g(x)$. In this case, $\lim_{x \to 0} \sin(\frac{1}{x})$ doesn't exist and the mentioned theorem isn't applicable. We can use the squeeze theorem:
$$|\sin(\frac{1}{x})|\le 1 \to -1\le\sin(\frac{1}{x})\le 1 \to -x\le x\sin(\frac{1}{x})\le x,\ \ \ x\ge 0$$
$$|\sin(\frac{1}{x})|\le 1 \to -1\le \sin(\frac{1}{x})\le 1 \to x\le x\sin(\frac{1}{x})\le -x,\ \ \ x\lt 0$$
Or using absolute value $$|\sin(\frac{1}{x})|\le 1 \to |x||\sin(\frac{1}{x})| = |x\sin(\frac{1}{x})| \le |x| \to -|x|\le x\sin(\frac{1}{x}) \le |x|$$
Using the fact $\lim_{x \to 0} |x| = \lim_{x \to 0} -|x| = 0$, we have $$\lim_{x \to 0} x\sin(\frac{1}{x}) = 0$$We can see this inequality clearly by plotting $x\sin(\frac{1}{x})$, $|x|$ and $-|x|$:
Looking at the plot, it seems that $x\sin(\frac{1}{x})$ oscillates forever but that's not true if we zoom out: 
This behavior can be explained by the limit $$\lim_{x \to \infty} x\sin(\frac{1}{x}) = 1$$
$\lim_{x\to 0}\sin(\frac{1}{x})$ is not infinity!
The sine function is an oscillating function that can take any value between $-1$ and $1$. The value approaching infinity is actually $\frac{1}{x}$ so think of it like the sine of a very large value. Since the sine function's range is $[-1,1]$ it is restricted to those values only, regardless of the input. So in reality the limit of $\sin(\frac{1}{x})$ as $x$ approaches $0$ doesnt exist, since all we know is that it belongs between $-1$ and $1$. So multiplying a value between $-1$ and $1$ with a value approaching $0$, you will have $0$
Hence the final limiting value will be $0$.
Suppose, $\lim_{x\to c}f(x) =L$ and $\lim_{x\to c}g(x) =M \space \space (L,M\in \overline{\Bbb{R}})$, then
$$\begin{align}\lim_{x\to c}[f(x) \cdot g(x)]&= \lim_{x\to c}f(x)\cdot \lim_{x\to c}g(x) = L\cdot M\end{align}$$
Observation:
$\lim_{x\to 0}x \cdot \frac{1}{x} \neq \lim_{x\to 0} x \cdot \lim_{x\to 0}\frac{1}{x} $
Conclusion: If both $\lim_{x\to c}f(x)$ and $\lim_{x\to c}g(x) $ exists then we can apply product rule of limit.
In your question,
$\begin{aligned}\lim_{x\to 0} x\sin\left(\frac{1}{x}\right)&=\lim_{x\to 0}x\cdot\lim_{x\to 0}\sin\left(\frac{1}{x}\right)\\&=0\cdot \infty\\&=\infty \end{aligned}$
$\lim_{x\to 0}\sin\left(\frac{1}{x}\right)$ doesn't exists.
$\lim_{x\to c} f(x) =L$ iff $\forall (x_n)_{n\in \Bbb{N}}$ with $(x_n) \to c\implies f(x_n) \to L$
$\lim_{x\to c} f(x) $ doesn't exists if $\exists (x_n), (y_n) $ such that $(x_n) \to c$ and $(y_n) \to c$ but
$\lim f(x_n) \neq \lim f(y_n) $
Now, choose $x_n = \frac{1}{n\pi}$ and $y_n= \frac{2}{(4n+1)\pi}$
Both $(x_n) , (y_n) $ converges to $0$ but
$ f(x_n) =\sin(n\pi) =0$
$f(y_n) =\sin(\frac{(4n+1)\pi}{2}) =1$
Hence, $\lim f(x_n)=0\neq 1 = \lim f(y_n) $
Hence, $\lim_{x\to 0} \sin(\frac{1}{x})$ doesn't exists.
