$\lim_{x \to c}f(x)=L>0, \lim_{x \to c} g(x)=\infty \Rightarrow \lim_{x \to c} f(x)g(x)=\infty$ $\lim_{x \to c}f(x)=L>0, \lim_{x \to c} g(x)=\infty \Rightarrow \lim_{x \to c} f(x)g(x)=\infty$
My attempt: Given $\varepsilon>0, \exists \delta_1>0$ such that $|x-c|<\delta_1 \Rightarrow |f(x)-L|<\varepsilon$ and $\exists \delta_2>0$ such that $|x-c|<\delta_2 \Rightarrow g(x)> \varepsilon$
I was trying to use the $L>0$, so I chose $\varepsilon=\dfrac L2$, which gave me $\dfrac L2 < f(x) < \dfrac {3L}{2}$
Again, $g(x)> \dfrac L2$, so $f(x)g(x)>\dfrac{l^2}{4}$
This gave me one particular choice of $\varepsilon$ for $f(x)g(x)$,but I think I need to prove that there is a $\delta_3>0$ such that $|x-c|<\delta_3 \Rightarrow f(x)g(x)>\varepsilon$, for any $\varepsilon \in \mathbb R$ right?
 A: You'd better use the two assumptions independently.
As you have shown, there exists $\delta_1>0$ such that $|x-c|<\delta_1$ , we have $\dfrac L2 < f(x) < \dfrac {3L}{2}$
Given $\epsilon>0$, since $L>0$, there exists $\delta_2>0$ such that for $|x-c|<\delta_2$, we have $ g(x)> \frac{\epsilon}{\frac{L}{2}}$
Hence let $\delta_3=\min\{\delta_1,\delta_2\}>0$, then for all $|x-c|<\delta_3$,
$$f(x)g(x)>\frac{L}{2}\frac{\epsilon}{\frac{L}{2}}=\epsilon$$
Hence we are done.
A: I think your problem stems from the use of symbolism ($\epsilon - \delta$ stuff). I will first provide the argument in plain English and then add $\epsilon, \delta$ later.
Since $f(x) \to L > 0$ as $x \to c$, it follows that $f(x)$ is near to $L > 0$ and therefore positive and away from $0$ as $x \to c$. And $g(x) \to \infty$ as $x \to c$ implies that $g(x)$ will exceed any given bound if we get too close to $c$. Multiplying $f(x)$ and $g(x)$ and noting that $f(x)$ is positive and away from $0$, it is easily seen that $f(x)g(x)$ will also exceed any given bound as $x \to c$.
Now the translation in $\epsilon, \delta$. Since $f(x) \to L$ as $x \to c$, we can take $\epsilon = L/2 > 0$ and note that there is a $\delta_{1} > 0$ such that $$L - \epsilon < f(x) < L + \epsilon$$ whenever $0 < |x - c| < \delta_{1}$. Thus we have $f(x) > L/2$ whenever $0 < |x - c| < \delta_{1}$. This is the translation for "$f(x)$ is positive and away from $0$ when $x$ is near $c$".
In order to show that $f(x)g(x) \to \infty$ as $x \to c$, we must be able to choose a $\delta > 0$ for a given $M > 0$ such that $f(x)g(x) > M$ whenever $0 < |x - c| < \delta$. Now we already know that $f(x) > L/2$ and hence to make $f(x)g(x) > M$ it should be enough to have $g(x) > 2M/L$. Since $g(x) \to \infty$ as $x \to c$, it is possible to choose $\delta_{2} > 0$ such that $g(x) > 2M/L$ whenever $0 < |x - c| < \delta_{2}$.
If we choose $\delta$ as the minimum of $\delta_{1}$ and $\delta_{2}$ and keep $0 < |x - c| < \delta$ then it will mean that both $0 < |x - c| < \delta_{1}$ and $0 < |x - c| < \delta_{2}$ are satisfied. Hence we will have both $f(x) > L/2$ and $g(x) > 2M/L$ whenever $0 < |x - c| < \delta$. This clearly shows that $f(x)g(x) > M$ whenever $0 < |x - c| < \delta$. This is how we choose $\delta$ based on $M$ and establish that $f(x)g(x) \to \infty$ as $x \to c$.
It has to be understood that this $\epsilon, \delta$ argument conveys the same meaning as given in my earlier informal argument and nothing more. Hence it is important to formulate the informal argument in one's mind to tackle the problem and then do an easy translation to $\epsilon, \delta$ type answer.
