# What's the best explanation of the fallacy in this 'paradox'?

Of course whenever you have two statements that each on its own sound plausible but then contradict each other, you can simply check which one is false by e.g. drawing a picture. But I hope that someone can find a clear explanation that addresses the heart of the misconception. So here is the paradox.

We know (for $$x \to +\infty)$$) that

$$e^{-x}$$ goes to zero faster than the reciprocal of any polynomial. $$\tag{1}$$

In particular we have that $$-e^{-x}$$ (which happens to be the derivative of $$e^{-x}$$) increases towards 0 faster than $$-\frac{1}{(x+1)^2}$$ (which happens to be the derivative of $$\frac{1}{x + 1}$$).

This means that the graph of $$e^{-x}$$ becomes flatter and flatter faster than that of $$\frac{1}{x + 1}$$ and since both start at $$(0, 1)$$ and tend to zero we find that $$e^{-x} \geq \frac{1}{x + 1}$$ and hence in particular:

$$\frac{1}{x + 1}$$ goes to zero faster than $$e^{-x}$$, $$\tag{2}$$

• Why do you even suppose that "become flatter faster" should imply "aproach the limit slower"?? Something like $\max\{0,1-100x\}$ becomes as flat as can get after $x=\frac1{100}$. -- In fac,t "become flatter faster" seems rather to suggest intuitively the opposite: "can take it easy because the goal is already almost reached" Oct 2 at 9:58
• Just to answer the first question 'Why do you even suppose that "become flatter faster" should imply "approach the limit slower"?': what happens is that I (sort of equate) 'become flatter faster' with 'being flatter' because most of the time this will be true. Oct 2 at 10:00
• We have : $-e^{-x}\geq -\frac{1}{(x+1)^2}$ integrating gives Oct 2 at 10:31
• What is the definition of "goes to zero faster"? Oct 2 at 11:31
• The behavior of a general function near $x=0$ has no relation whatsoever with what happens for $x \to \infty$ - so you cannot apply arguments related to the behavior at $\infty$ near $0$ Oct 3 at 6:09

You got me for a minute until I plotted these on Desmos.

As you can see, $$-\frac{1}{(x+1)^2}$$ initially increases faster than $$-e^{-x}$$. Then, it was overtaken at around $$x=2.513$$.

This is enough to explain your "paradox", because $$e^{-x}$$ initially decreases faster than $$\frac{1}{x+1}$$.

In general, asymptotics only apply "for large enough $$x$$" and therefore should not generate paradoxes like yours. • I like all the answers, but I am accepting this one, based on the popular vote, and also I think in this case a picture is the most powerfull tool to understand what's going on Oct 8 at 13:33

I have re-written without the Paradox :

Claim X : Curve $$C1 :: e^{-x}$$ goes to zero faster than the reciprocal of any polynomial. [ true ]

In particular we have that $$-e^{-x}$$ (which happens to be the derivative of $$e^{-x}$$) increases towards 0 faster than $$-\frac{1}{(x+1)^2}$$ (which happens to be the derivative of $$\frac{1}{x + 1}$$). [ true ]

This means that the graph of $$e^{-x}$$ becomes flatter and flatter faster than that of Curve $$C2 :: \frac{1}{x + 1}$$ ... [ true ]

... and since both start at $$(0, 1)$$ and tend to zero we find that $$e^{-x} \geq \frac{1}{x + 1}$$ ... [ not true ]
... and since both start at $$(0, 1)$$ and tend to zero we find that $$e^{-x} \leq \frac{1}{x + 1}$$ ... [ true ]

... and hence in particular:

... $$\frac{1}{x + 1}$$ goes to zero faster than $$e^{-x}$$ [ not true ]
... $$e^{-x}$$ goes to zero faster than $$\frac{1}{x + 1}$$ [ true ]

... contradicting Claim X. [ not true ]
... which is consistent with Claim X. [ true ]

Elaboration :

When a curve becomes flatter , that means it is "reaching" the "Destination" limit.
When a curve becomes flatter faster , that means it is "reaching" the "Destination" limit faster.
When Curve $$C1$$ is "reaching" the "Destination" faster than Curve $$C2$$ , then Curve $$C1$$ must be closer to the Destination than Curve $$C2$$.

Destination in this Case is $$0$$ , hence [[ $$e^{-x}$$ is closer to $$0$$ ]] more than [[ $$\frac{1}{x+1}$$ is closer to $$0$$ ]] !

Putting that in Math Syntax , we will get necessary Conclusion without Paradox !

This viewpoint is at least implicit in the existing comments and answers, but maybe can be boiled down further:

Suppose Car 1 and Car 2 start at point $$A$$ at time $$0$$ and end at point $$B$$ at $$t = 1$$. Eventually Car 1 is slower than Car 2. Which car is ahead at the end of the race (while Car 1 is slower)?

If that doesn't clarify the intuition, run time backward from $$t = 1$$: Two cars start at $$B$$ at the same time. Which car is initially behind (i.e., closer to $$B$$), the slower Car 1, or the faster Car 2?

(Technical note: Using $$t = 1$$ versus letting $$t \to \infty$$ is immaterial. The closed unit interval $$[0, 1]$$ and the extended positive reals $$[0, \infty]$$ are equivalent as ordered sets.)

• I really like this. By translating everything to something that happens in finite time, we are back in the world of things we can imagine. Curiously it is also possible to make a similar car-analogy for the incorrect intuition (that might even help some readers see what I was trying to say). I'll put it in the next comment. I think putting the two together really sheds some light on what is going on. Oct 2 at 21:06
• Alternative version: Suppose Car 1 and Car 2 start at point $A$ at time $0$ and end at point $B$ at times $t = T_1$ and $t = T_2$ respectively. For most of the time, Car 1 is slower than Car 2. Who arrives at $B$ first? (Of course the answer is that we have too little information to know, but intuition is strongly pulling us towards Car 2, I'd say) Oct 2 at 21:08
• Your second comment may be a good example of how expectations from qualitative English ("cars all travel about the same speed") and consequences of mathematical hypotheses can diverge widely. Hagen von Eitzen's comment, for example, shows the winning car can have positive speed for an arbitrarily short time by arriving at $B$ almost instantaneously, then staying put. <> I'd be more inclined to call the second comment a mis-directed perception than a paradox, however. :) Oct 3 at 0:51