The specific problem with regards to this graph is:
$$\lim_{x \to -\infty} g(2 + e^x)$$
Now it's true that we technically can't apply "the limit of a sum is the sum of the limits" since the function is not continuous. But to me, this intuitively seemed like the solution would just be:
$$ g(2 + 0) $$
$$ = g(2) $$
$$ = 0.5 $$
Admittedly, this seems grounded on foolish assumptions, in hindsight. But even if my approach isn't mathematically sound, I don't get why the provided approach on the answer key should be sound:
$$ u = 2 + e^x $$
$$ as\ x\to – \infty, u\to2^+ $$
$$ \lim_{u \to 2+} g(u) = -2 $$
I understand that, when you look at the graph of $e^x$ as it approaches negative infinity, it gradually diminishes towards zero, so I somewhat understand how the value it approaches is sort of like approaching $0$ from the right ($0^+$) — but to me, this provided solution looks like it's defying the fact that $ \lim_{x \to -\infty} e^x = 0 $, and it seems really alien to me.
One thing I'm wondering is that things are different since we're working inside of the function $g$, but I'm unable to find any such rule or explanation for this.
Best Answer
You're right: $e^x$ tends to $0$ as $x \rightarrow - \infty$. But, note that $e^x$ takes only positive values, so $e^x$ tends to $0$ from above (aka $e^x \rightarrow 0^+$). This is a stronger fact than $e^x \rightarrow 0$, in that the former imlies the latter, but not the other way around.
As for citing a theorem that proves this, it's often tricky, as there's rarely a theorem stated in the various text books that deals with all kinds of one-sided limits. It's tricky, as there are so many varieties of limits (two-sided, each one-sided type, $\pm \infty$) to state it in a single theorem (one can elegantly state it with the notion, from topology, of a "direction", which unifies the types of limits).
But yes, in this case, it should be intuitive enough to see that $2 + e^x \rightarrow 2^+$, as $2 + e^x \rightarrow 2$ and $2 + e^x > 2$ for all $x$. Also intuitively, $\lim_{x \rightarrow -\infty}g(2 + e^x) = \lim_{y \rightarrow 2^+}g(y)$. This all can be proven using $\varepsilon$-$\delta$ limit definitions of limits too, but until you learn them, it's going to be difficult to properly justify statements to do with limits.