# How can e^x seemingly approach 0+ as x approaches negative infinity? 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.

• It appears (from this question as well as your previous one dealing with $u=x^{2}$) that you are trying to think too deeply in these matters (one sided limits). When $x\to - \infty$ it is well known fact that $e^{x} \to 0$ (proof is not trivial btw) and it is also known that $e^x$ is always positive hence $u=2+e^{x}\to 2^{+}$ and the desired answer is $-2$. – Paramanand Singh Sep 4 '17 at 5:50
• About your last sentence (working inside $g$), you should note that $g$ has no role to play in substitution $u=2+e^{x}$. $g$ comes into picture after the substitution and not before. – Paramanand Singh Sep 4 '17 at 5:53
• @ParamanandSingh Perhaps I am looking too deeply into things, but I work better when I understand why something is so -- in any case, as you recall from my last question, the answer key can be wrong, so I'm skeptical of it. – K_7 Sep 6 '17 at 4:48
• I appreciate your focus on understanding (contrary to majority who equate maths with rote problem solving techniques). But in this case you are perhaps trying to make a trivial thing as non-trivial. Most topics in calculus are not as difficult as the cheap textbooks have made them. Basic laws of limits (including the $\epsilon, \delta$ gymnastics) are the easier parts while the theorems based on completeness of real numbers (like intermediate value theorem or mean value theorem) are harder to understand. And yes one has to be in general skeptical when studying calculus textbooks. – Paramanand Singh Sep 6 '17 at 4:58

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.
• @K_7 Good catch! I've changed the $\infty$ to a $-\infty$. – Theo Bendit Sep 4 '17 at 4:22
• The first line is missing the minus too in $x\to\color{red}{-}\infty$. – A.Γ. Sep 4 '17 at 5:19