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While in Calculus the other day I stumbled upon a contradiction in L'Hopitals Rule vs. Special Trig Limits.

The problem looks like this

$$ \lim_{x\to 0} \frac{\tan(x)−x}{x^3} $$

Using L'Hopitals rule (because the limit = 0/0) $$ \lim_{x\to 0} \frac{\tan(x)−x}{x^3} = \lim_{x\to 0} \frac{\frac{d}{dx}(\tan(x)−x)}{\frac{d}{dx}(x^3)} = \lim_{x\to 0} \frac{\sec(x)^2−1}{3x^2} $$ Again using L'Hopitals rule (because the limit = 0/0) $$ \lim_{x\to 0} \frac{\sec(x)^2−1}{3x^2} = \lim_{x\to 0} \frac{\frac{d}{dx}\sec(x)^2−1}{\frac{d}{dx}3x^2} = \lim_{x\to 0} \frac{2\sec(x)^2\tan(x)}{6x} = \lim_{x\to 0} \frac{\sec(x)^2\tan(x)}{3x} $$ Again using L'Hopitals rule (because the limit = 0/0) $$ \lim_{x\to 0} \frac{\sec(x)^2\tan(x)}{3x} = \lim_{x\to 0} \frac{\frac{d}{dx}\sec(x)^2\tan(x)}{\frac{d}{dx}3x} = \lim_{x\to 0} \frac{4\sec(x)^2\tan(x)^2 + 2\sec(x)^4}{6} = \frac{1}{3} $$ Now, using special trig limits... $$ \lim_{x\to 0} \frac{\tan(x)−x}{x^3} = \lim_{x\to 0} \frac{\tan(x)}{x^3}-\frac{x}{x^3} = \lim_{x\to 0} \frac{\tan(x)}{x}*\frac{1}{x^2}-\frac{1}{x^2} $$ With knowledge of $\tan(x)$ trig limit $$ \lim_{x\to 0} \frac{\tan(x)}{x} = 1 $$ You can now simplify. $$ (\lim_{x\to 0} \frac{\tan(x)}{x}*\frac{1}{x^2}-\frac{1}{x^2}) = (\lim_{x\to 0} \frac{\tan(x)}{x}*\lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2}) = (1 * \lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2}) $$ Now the end result $$ \lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2} = \lim_{x\to 0}\frac{1}{x^2}-\frac{1}{x^2} = 0 $$ 1/3 does not = 0. Therefore there is a discrepancy through special trig limits and L'Hopitals rule. There is the same anomaly with $$ \lim_{x\to0}\frac{\sin(x)-x}{x^3} = (\text{through L'Hopitals}) -\frac{1}{6} \text{ or (through special trig limits) } 0 $$ On a calculator in graph or table mode as you approach 0 it seems to be 1/3. But as it turns out, if you get very precise, the limit actually seems to be approaching 0.

This happens the same with the graph, it seems to be parabolically approaching 1/3 but if you zoom in to an extreme you see it actually approaches zero.

This original work is my own as published on Sunday, October 9, 2016 at 9:51pm. I claim all knowledge credits and fallacies that may come with this discrepancy. But overall please, prove or disprove this, or at least explain why this occurs. Thank you for your time.

(I will try to get a picture of the graph and table)

Jean-Claude Arbaut
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kjhurley99
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    $\frac{1}{3}$ is the correct answer. The issue with your approach is that you evaluated $\lim_{x\to 0}\frac{\tan x}{x}$ before evaluating the rest of the limit. – carmichael561 Oct 10 '16 at 02:16
  • http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhôpital-rule-or-series-expansion – lab bhattacharjee Oct 10 '16 at 03:09
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    Did your teacher really say he had never seen anything like this before? Such mistakes like the one in your question are very common with beginners in calculus and are normally a result of using the limit laws casually without paying attention to the conditions under which those limit laws hold. Also the usage of L'Hospital many many times to evaluate the limit $(\tan x - x)/x^{3}$ is very convoluted. Just use LHR once to get $(\sec^{2}x - 1)/3x^{2} = (1/3)(\tan^{2}x/x^{2})$ and this tends to $1/3$. – Paramanand Singh Oct 10 '16 at 06:59
  • @ParamanandSingh I realize I had emphasized the wrong part of this question. Its was more of the calculator question, I'll revise in a little to make it make more sense. Thank you! – kjhurley99 Oct 11 '16 at 02:40
  • So much simpler when you use Taylor's formula for the tangent... $$\tan x=x+\frac{x^3}{3}+O(x^5)$$ – Jean-Claude Arbaut Oct 13 '16 at 06:35

2 Answers2

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Your error occurs at this step,

"You can now simplify

$(\lim_{x\to 0} \frac{tan(x)}{x}*\frac{1}{x^2}-\frac{1}{x^2}) = (\lim_{x\to 0} \frac{tan(x)}{x}*\lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2}) = (1 * \lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2})$

In particular, this equality

$(\lim_{x\to 0} \frac{tan(x)}{x}*\frac{1}{x^2} = \lim_{x\to 0} \frac{tan(x)}{x}*\lim_{x\to 0}\frac{1}{x^2}$.

You can only split the products when both limits are finite. $\lim_{x\to 0}\frac{1}{x^2} = \infty$

G. Snapsmath
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  • Ooh I see. Is that included in the common Limit Multiplication Law, and can you prove that to be true? Also, why does the calculator have a discrepancy as you get closer to 0 especially with the graph? – kjhurley99 Oct 10 '16 at 02:29
  • You can pick up any analysis book and see the proof of limit multiplication(I suggest Rudin). I will give an explicit example of what happens when one limit is infinity. Consider $\lim_{x\to\infty} \frac{x}{x}$. Clearly by first reducing this limit is 1. However, let's see what happens when we split it. $\lim_{x\to\infty} x \cdot \lim_{x\to\infty} \frac{1}{x} = 0\cdot \infty$ – G. Snapsmath Oct 10 '16 at 02:33
  • Thank you for the clarification. Then do you think the calculator (a Ti-89) is starting to "give up" once you approach a certain small value, or rather its infintesimally small occilation past a certain range gives it a quality of both 0 and 1/3? Because it does say it approaches zero to around the x --> 1e-8 range – kjhurley99 Oct 10 '16 at 03:26
  • What did you put in the calculator? – G. Snapsmath Oct 10 '16 at 03:29
  • (tan(x)-x)/x^3 you need to zoom in a lot on x = 0 for graph or use table – kjhurley99 Oct 10 '16 at 03:34
  • It is odd, but I am not getting the same thing. – G. Snapsmath Oct 10 '16 at 03:38
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The error you make can be reproduced in a simpler context: let $f(x)=x+x^2$; then $$ \lim_{x\to0}\frac{f(x)}{x}=1 $$ With the same argument you used, we should have $$ \lim_{x\to0}\frac{f(x)-x}{x^2}= \lim_{x\to0}\left(\frac{f(x)}{x}\cdot\frac{1}{x}-\frac{1}{x}\right) $$ and, since $\lim_{x\to0}f(x)/x=1$ we should get $0$. Wrong!

Indeed $$ \lim_{x\to0}\frac{f(x)-x}{x^2}=\lim_{x\to0}\frac{x^2}{x^2}=1 $$

There are many similar cases: $$ \lim_{x\to0}\frac{x-\sin x}{x^3}= \lim_{x\to0}\left(\frac{1}{x^2}-\frac{\sin x}{x}\frac{1}{x^2}\right) $$ does not allow to substitute $1$ for $\frac{\sin x}{x}$.

As you can see, this has nothing to do with l'Hôpital's rule.

You can “remove” a global factor having limit $1$ without modifying the limit; for instance, $$ \lim_{x\to0}\frac{\sin(\sin x)-\sin x}{x^3}= \lim_{x\to0}\frac{\sin(\sin x)-\sin x}{\sin^3x} \underbrace{\frac{\sin^3 x}{x^3\mathstrut}}_{(*)}= \lim_{x\to0}\frac{\sin(\sin x)-\sin x}{\sin^3x} $$ because the part marked with $(*)$ has limit $1$. More precisely, the limit we started with exists if and only if the final one exists and, in this case, they are equal.

You may not remove such a factor if it only multiplies a summand in the whole thing, just like we can't remove $\frac{f(x)}{x}$ from the example limit above. And you should be able to see the reason why. If $\frac{f(x)}{x}$ was a “global multiplier”, there would be no problem in removing it. But $f(x)-x=x^2$, not zero. And this remainder is what counts in $$ \lim_{x\to0}\frac{f(x)-x}{x^2} $$ exactly like $\tan x-x$ is what counts in $$ \lim_{x\to0}\frac{\tan x-x}{x^3} $$

By the way, if you remember that $$ D\tan x=\frac{1}{\cos^2x}=1+\tan^2x $$ your application of l'Hôpital would end after the first step: $$ \lim_{x\to0}\frac{\tan x-x}{x^3}= \lim_{x\to0}\frac{1+\tan^2x-1}{3x^2}= \lim_{x\to0}\frac{1}{3}\left(\frac{\tan x}{x}\right)^{\!2} $$

egreg
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