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Assuming that $\lim_{x \rightarrow a} f(x) = L$ where $L \neq 0$, and $\lim_{x \rightarrow a} g(x)$ does not exist, is it true that $\lim_{x \rightarrow a} [f(x)*g(x)]$ does not exist?

This is to be proved using the laws of limits.

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  • $\begingroup$ I'm pretty sure that is the only possibility $\endgroup$ – graydad Sep 4 '14 at 4:55
  • $\begingroup$ Suppose limit of $fg$ exists. Then $g=fg/f$ so that limit of $g$ also exists. This contradiction shows that limit of $fg$ can't exist. Note that $L\neq 0$ is a must for this argument to work. $\endgroup$ – Paramanand Singh Sep 4 '14 at 13:42
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For the sake of contradiction, suppose $\lim_{x \to a}[f(x)g(x)]=L'$ for some finite L'. By limit laws we know $$\lim_{x \to a}[f(x)g(x)]=[\lim_{x \to a}f(x)]*[\lim_{x \to a}g(x)]=L'$$ We may go one step further and say $$[\lim_{x \to a}f(x)]*[\lim_{x \to a}g(x)]=L*[\lim_{x \to a}g(x)]=L'$$ Now divide by $L$ on both sides and we have $$[\lim_{x \to a}g(x)]=\frac{L'}{L}$$ Since $L\neq0$ and both $L$ and $L'$ are finite that means $\frac{L'}{L}$ is finite. This contradicts the assumption that the limit of $g(x)$ does not exist.

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  • $\begingroup$ This gives more idea than my answer... Good one!!! $\endgroup$ – user87543 Sep 4 '14 at 5:08
  • $\begingroup$ @PraphullaKoushik Thank you :) $\endgroup$ – graydad Sep 4 '14 at 5:12
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    $\begingroup$ I think you should rephrase this argument using the formula for the limit of a quotient. The limit of a product formula requires that you know the individual terms have limits, but you're told specifically that $\lim_{x\to a}g(x)$ doesn't exist. $\endgroup$ – Kim Jong Un Sep 4 '14 at 5:41
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For $f\equiv1$ and $g(x)=\frac{1}{x}$ we have :$$\lim_{x\rightarrow 0} f(x)=1$$$$\lim_{x\rightarrow 0}g(x)=\infty$$ $$\lim_{x\rightarrow 0}f(x)g(x)=??$$

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