Prove the following using epsilon-delta definition:
Let $a \in \mathbb{R}$ and $f,g$ be defined on $\mathbb{R}$.
If $$\lim_{x\to a} f(x) = 647 \tag 1$$ and $$\lim_{x\to a} g(x) = \infty \tag 2$$ then
$$\lim_{x\to a} f(x) + g(x) = \infty.$$
$$(1) \implies\forall \epsilon > 0, \exists \delta > 0, s.t. |x-a|<\delta \implies |f(x)-647| < \epsilon$$
$$(2) \implies \forall M > 0, \exists \delta_2 > 0, s.t. |x-a|<\delta_2 \implies |g(x)| > M$$
Not sure about this... I've set up the definitions. I believe we will have to take a minimum between the deltas for first two limits then use it in the third.
Anyone have any clues?
Let $M > 0$. You want to find a $\delta'>0$ such that $|x-a| < \delta'$ implies $f(x) + g(x) \ge M$.
By $(1)$ there exists $\delta$ such that $|x-a| < \delta$ implies $|f(x) - 647| < 1$, which implies $f(x) > 646$.
By $(2)$ there exists $\delta_2$ such that $|x-a| < \delta_2$ implies $g(x) > M$.
Then with $\delta' = \min\{\delta, \delta_2\}$, we have $|x-a| < \delta'$ implies both $f(x) > 646$ and $g(x) > M$, so we have $f(x) + g(x) > 646 + M > M$.