# ε-δ definition and M-δ definition to prove M-δ limit

(i)$\lim_{x \to a} f(x)= L$ (ii)$\lim_{x \to a} g(x)= \infty$

Using the definitions of $\varepsilon-\delta$ definition and $M-\delta$ definition to prove $lim_{x \to a} f(x)g(x)=\infty$ as $x$ approaches $a$.

Anyone can help me with this? I have done my definition part respectively but don really know how to continue to combine those definitions to prove that.

I was given this solution:But I could not understand, can anyone explain to me please?

let $\varepsilon=\frac{L}{2}$ in part(i), there exists $\delta_1 >0$ such that $0<|x-a|< \delta_1$,then $|f(x)-L|<\frac{L}{2}$ for all $M>0$ In part(ii),there exists $\delta_2>0$ such that $g(x)>\frac{2M}{L}$ for $0<|x-a|<\delta_2$ Then for all $M>0$,there exists $\delta=\min(\delta_1,\delta_2)$ such that $0<|x-a|<\delta$ Then we have $f(x)g(x) \geq min(L-\frac{L}{2},L+\frac{L}{2})g(x)=0.5 \cdot L \cdot g(x)>(\frac{L}{2})*(\frac{2M}{L})=M$

I understand that it wants to show $f(x)g(x)>M$ because of definition. but I don understand what is going on(especially the last sentence where ...$min(L-\frac{L}{2},L+\frac{L}{2})$, can anyone explain to me the approach for solving this ques?thx!

• Consider $L=a=0$, $f(x) = x$, $g(x) = |\log(x)|$, then the conditions are satisfied but $\lim\limits_{x \to 0} f(x)g(x) = 0$. – Nigel Overmars Nov 30 '14 at 16:51

You want to assume $L > 0$.
We are given $\lim_{x \to a } f(x) = L$. By defintion, this means that for all $\epsilon > 0$, we can always find some $\delta_1 > 0$ such that if $0 < |x-a| < \delta_1$ , then $|f(x) - L | < \epsilon$. Since this holds for any positive $\epsilon$, we can choose $\epsilon = \frac{L}{2} > 0$. In particular, $f(x) > \frac{L}{2}$. (PRoof?)
Next, since $\lim_{x \to a} g(x) = \infty$, then by definition for any $\alpha > 0$, we can always find some $\delta_2 > 0$ such that if $0 < |x-a| < \delta_2$, then $g(x) > \frac{2\alpha}{L}$. Now, we want to show that $\lim_{x\to a } f(x)g(x) = \infty$. Choose $\delta = \min( \delta_1, \delta_2 )$. Then, if $0 < |x-a| < \delta$, we must have
$$fg > \frac{L}{2} \frac{2 \alpha}{L} = \alpha$$