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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

For $f\equiv1$ and $g(x)=\frac{1}{x}$ we have :$$\lim_{x\rightarrow 0} f(x)=1$$$$\lim_{x\rightarrow 0}g(x)=\infty$$ $$\lim_{x\rightarrow 0}f(x)g(x)=??$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.