Proving $\lim_{x \to 0}f(x)g(x) = c$ for any number c when one function tends to infinity and the other to zero

Question

I'm trying to attempt one of the problems in a Calculus book and this is the question:

Show that for any number $c$, it is possible to find functions $f$ and $g$ such that $\lim_{x \to 0} f(x) = \infty$, $\lim_{x \to 0} g(x) = 0$, and $\lim_{x \to 0}f(x)g(x) = c$

From the question, I understand we have to prove that $\lim_{x \to 0}f(x)g(x) = c$ for any number $c$ which means that the limit is undefined. So, I'm aiming to prove that $0 < |x| < \delta \implies |f(x)g(x) - c| > \epsilon$

After making the assumption of $\delta = min(\delta_1, \delta_2)$ (The other deltas correspond to the ones in the limit definition of f(x) and g(x)), I'm able to infer that $f(x) > \epsilon$ and $g(x) < \epsilon$. But I'm not able to proceed from here. How to prove the above thing ?

Answer 1

You are only asked to show that it is possible to find such functions.

You can verify that $f(x)=\dfrac{1}{|x|}, g(x)=c|x|$ satisfy the requirements.

Answer 2

What you should prove is that, given $c$, there are function $f$ and $g$ such that

• $\displaystyle\lim_{x\to0}f(x)=\infty$;
• $\displaystyle\lim_{x\to0}g(x)=0$;
• $\displaystyle\lim_{x\to0}f(x)g(x)=c$.

Just take $f(x)=\dfrac c{|x|}$ and $g(x)=|x|$, unless $c=0$. In that case, take $f(x)=\dfrac1{x^2}$ and $g(x)=x^3$, for instance.