I wanted to prove this: $\lim_{x\rightarrow c} f(x)g(x)= \infty$ when $\lim_{x\rightarrow c} f(x) = L (L>0)$ and $\lim_{x\rightarrow c} g(x) = \infty$ by using epsilon-N method.
I used two inequalities derived by using epsilon-delta and epsilon-N method on two limit conditions. I thought that by fixing L as $L>\epsilon>0$, I could make the form of $f(x)g(x)>K (K>0)$ using inequalities. Well I cannot derive that form, my result was always $K<0$. How can I prove this statement? Thank you.
Let $M>0$. Since $f(x)\to L$ id $x\to c$, there is $\delta_1 >0$ s.t. $$f(x)\geq \frac{L}{2}>0,$$ whenever $|x-c|<\delta_1 $. Moreover, since $g(x)\to \infty $, when $x\to c$, there is $0<\delta <\delta _1$ s.t. $$g(x)\geq \frac{2M}{L},$$ if $|x-c|<\delta $. In particular, if $|x-c|<\delta $, we have $$f(x)g(x)\geq M,$$ and thus $f(x)g(x)\to \infty $ whenever $x\to c$.