# 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?

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.

• Sorry for the confusion. I have edited my title. I am asking 'the limit of product of two functions'. – Math learner Oct 15 '19 at 0:37
• If $f(x)$ tends to a non-zero then the limiting behavior of $f(x) g(x)$ is same as that of $g(x)$. – Paramanand Singh Oct 15 '19 at 8:32

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.
• Can you explain why $\lim_{x \to x_0}f(x)g(x) = M$ then you get the limit of g(x). it looks like you are using what you want to prove. – Math learner Oct 15 '19 at 0:35
• Assume that the limit $f(x)g(x)$ exists, by way of contradiction as you did. Then we just use the fact that the limit of a quotient is the quotient of the limits. – saulspatz Oct 15 '19 at 0:37
If $$f$$ has limit $$L\neq 0$$ then you can bound below $$|f(x)|$$ in some punctured neighbourhood of $$x_0$$. (For example, if $$L$$ is positive then you can get $$f(x)>L/2$$).