[Math] If one limit of function exists and does not equal to zero and the other does not have a limit, does the limit of product of two functions exist

Question

I suppose $\lim_{x\to x_0}f(x) = L$ and L $\neq 0$ and limit of g(x) does not exist. For this statement, I think the statement is true, and I want to prove it by contradiction.

Suppose $\lim_{x \to x_0}f(x)g(x) = M$. Based on the definition of limit, $\forall \epsilon > 0, \exists \delta > 0$ such that $\forall x: |x – x_0| < \delta$ we have $|f(x)g(x) – M| < \epsilon$. How could I play with absolute value and triangular equality so that I can show that the limit of g(x) exists and thus we have a contradiction.

Answer 1

If $\lim_{x\to x_0}f(x)=L\neq0$ and $\lim_{x\to x_0}f(x)g(x)=M$ then $$\lim_{x\to x_0}g(x) =\lim_{x\to x_0}{f(x)g(x)\over f(x)}={M\over L}$$ contradiction.