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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.

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

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  • $\begingroup$ 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. $\endgroup$ – Math learner Oct 15 at 0:35
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    $\begingroup$ 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. $\endgroup$ – saulspatz Oct 15 at 0:37
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Hint:

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 $f(x)>L/2$).

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