If $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = c$, prove that $\lim_{x \to a} f(x)g(x) = \infty$ 
How can we prove that $\lim_{x \to a} f(x)g(x) = \infty$ via $\varepsilon-\delta$, given that $\lim_{x \to a} f(x) = \infty$ and $\lim_{x \to a} g(x) = c$, where c is a positive number?

I have came up with a proof but am unsure if it is a correct approach.
$\lim_{x \to a} f(x)g(x) = \infty$
For $M>0$, there exists a proper $\delta>0$ such that $0<|x-a|<\delta => f(x)g(x)>M$
Since $\lim_{x \to a} f(x) = \infty$,
$0<|x-a|<\delta_{1} \implies f(x)>M_{1}$
Since $\lim_{x \to a} g(x) = c$,
$0<|x-a|<\delta_{2} \implies |g(x)-c|<\varepsilon$
$\therefore -\varepsilon + c < g(x) < \varepsilon + c$
$\therefore$ It suffices to take $M = M_{1}(\varepsilon + c)$
Proof:
Let $M > 0$, there exists a proper $\varepsilon > 0$ such that $0<|x-a|<\delta \implies f(x)g(x)>M$
Let $M_{1} = \frac{M}{\varepsilon + c}$, $\varepsilon > 0$, $c > 0$
$\therefore 0<|x-a|<\delta_{1} \implies f(x) > M_{1} = \frac{M}{\varepsilon + c}$
$\therefore 0<|x-a|<\delta_{2} \implies |g(x)-c| < \varepsilon => -\varepsilon + c < g(x) < \varepsilon + c$
Choose $\delta = \min\{\delta_{1}, \delta_{2}\}$
$\therefore 0<|x-a|<\delta \implies f(x)g(x) > \frac{M}{\varepsilon + c}\times(\varepsilon+c) = M$
 A: You have all the right ideas, and your proof could be considered correct, given some benefit of the doubt. But in early analysis courses, it's all about being completely precise, therefore I have here written a more streamlined proof, which explain every step meticulously.
We want to prove that $\lim_{x\rightarrow a} f(x)g(x) = \infty$. In other words, given a positive $M$, we want to find $\delta >0$ such that $f(x)g(x) > M$ for all  $x$ with $|x-a|<\delta$. Now that we have written down exactly what to prove, here is how I would prove it.
Let $M>0$. Since we want to show that the limit is positive infinity, let us show that $g(x)$ stays positive if x is close enough to $a$. Since $c>0$, we have $\varepsilon = \frac{c}{2}>0$. Therefore, we can choose a $\delta_1>0$ such that $f(x)>\frac{c}{2}>0$, given that $|x-a|<\delta_1$
Now let $N = \frac{M}{\frac{c}{2}}>0$. Since $\lim_{x \rightarrow a}f(x) = \infty$, we can choose $\delta_2>0$ such that $f(x)>N$ when $|x-a|<\delta_2$. Now it follows that if we set $\delta = \min(\delta_1, \delta_2)$, we get for $|x-a|<\delta$ that
$f(x)g(x)>N\frac{c}{2}$=M
as desired.
