$\lim_{x \to a} f(x)= L$ (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.

Appreciate your help!

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 δ1>0 such that $0<|x-a|< \delta_1$,then $|f(x)-L|<\frac{L}{2}forallM>0$ In part(ii),there exists δ2>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!


The result is not correct if $L=0.$ So, the first assumption must be $L\ne 0.$ Also, since in the proof it is shown "something is bigger than" it must be $L>0.$ Note also that $\epsilon=L/2>0$ implies $L>0.$ In such a case, we have

$$|f(x)-L|<\frac L2\implies L-f(x)<\frac L2\implies f(x)>\frac L2.$$ There is no need for considering $\min\{L-\frac{L}{2},L+\frac{L}{2}\},$ since $L+\frac{L}{2}>L-\frac{L}{2}.$

The idea of the proof is as follows: for $x-a$ small enough $f(x)$ is near $L,$ $g(x)$ is arbitrary large, and, thus, $f(x)$ is $"L\times$ arbitrary large=arbitrary large".

For $\epsilon=L/2$ there exists $\delta_1$ such that $$0<|x-a|<\delta_1\implies f(x)>L/2$$ and there exists $\delta_2$ such that $$0<|x-a|<\delta_2\implies g(x)>2M/L.$$ Then

$$0<|x-a|<\min\{\delta_1,\delta_2\}\implies f(x)g(x)>\frac L2\frac{2M}{L}=M.$$


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