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I was reading a Calculus book and I saw this problem which looks easy: $$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} = ?$$ It's a 0/0 limit and it's using some of the Taylor series of $\sin$ and $\cos$ expressions to solve the problem.

I know that the First and Second way should be correct because it's using more expressions of the Taylor series around 0. What I can't figure out is WHY using fewer expressions of the Taylor series in the Third way doesn't give 0/0 but gives a wrong answer?

First way: $$\lim _{x \rightarrow 0} \frac{2 x \cos x-2 \sin x \cos x}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x(x-\sin x)}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x\left(x-x+\frac{x^3}{6}\right)}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x\left(\frac{x^3}{6}\right)}{x^3}=\frac{1}{3}$$ Second way: $$\lim _{x \rightarrow 0} \frac{2x(1-\frac{x^2}{2})-(2x-\frac{8x^3}{6})}{x^3}=\lim _{x \rightarrow 0} \frac{2x-x^3-2x+\frac{8x^3}{6}}{x^3}=\lim _{x \rightarrow 0} \frac{\frac{x^3}{3}}{x^3}=\frac{1}{3}$$ Third way: $$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} =\lim _{x \rightarrow 0} \frac{2 x \cos x-2x}{x^3}=\lim _{x \rightarrow 0} \frac{2x(\cos x -1)}{x^3}=\lim _{x \rightarrow 0} \frac{2x(-\frac{x^2}{2})}{x^3}=-1$$

user21820
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Javad
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  • $2\sin 2x \approx 4x$ and not $2x$. – Gary Nov 01 '20 at 14:42
  • @Gary Yes, thank you. I'm gonna edit that. – Javad Nov 01 '20 at 14:43
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    The main thing is that, in general, you cannot just replace functions by their arbitrary Taylor polynomials and expect the limit to remain the same. You always have to inlude error terms, otherwise you can get wrong answers. In the first 2 cases you were lucky, and using polynomials did not affect the value of the limit. In the third case, the first equality is simply not true, you are changing to a different problem. – Gary Nov 01 '20 at 14:48
  • @Gary Thank you! But how can we inlude error terms? and how can we guarantee using some Taylor polynomials doesn't give wrong answers? – Javad Nov 01 '20 at 14:52
  • @Gary I can't figure out why in the third case, the first equality is not true. – Javad Nov 01 '20 at 14:56
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    @Javad A priori, none of the equalities where you substitute the trig functions for Taylor polynomials are true. Because the sine and cosine functions are not equal to their Taylor polynomials, in any finite degree. With the first two you were lucky to include enough terms to not change the limit, but your main takeaway from this really ought to be that that was luck, and not actually a valid manipulation. If you want to be certain that the equality remains true after such a substitution, you need to collect the error terms somewhere. – Arthur Nov 01 '20 at 15:06
  • @Arthur Thank you! I wasn't paying attention to the error terms, by the way, it was a nice answer :))) – Javad Nov 01 '20 at 15:11
  • Duplicate of: Is it okay to "ignore" small numbers in limits where $x$ approaches infinity? – user21820 Nov 02 '20 at 02:23

2 Answers2

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You are dividing by $x^3$ at the end, so you need all possible terms at least of degree $3$ in the numerator to be present, otherwise you're basically guaranteed to change the value of the limit.

Let's keep the error term in the third way, and see what happens. I will do that the following way: we have $$ \sin(2x) = 2x + x^3\cdot g(x)\\ \cos(x) = 1-\frac{x^2}2 + x^4\cdot h(x) $$ for some functions $g$ and $h$ where $g(x)$ and $h(x)$ are bounded as $x\to 0$. (It is more common to use $O(x^3)$ instead of $x^3\cdot g(x)$ and $O(x^4)$ rather than $x^4\cdot h(x)$. But the $O$ terms can be a bit unintuitive to arithmetize with, so if you are unaccustomed to working with error terms, I think that my approach here is closer to what you are already used to.)

Then we follow the steps in your third way and see what we get: $$ \frac{2 x \cos x- \sin 2x}{x^3} =\frac{2 x \cos x-(2x + x^3\cdot g(x))}{x^3}\\ =\frac{2x(\cos x -1) - x^3\cdot g(x)}{x^3}\\ =\frac{2x(-\frac{x^2}{2} + x^4\cdot h(x)) + x^3\cdot g(x)}{x^3}\\ =-1 + x\cdot h(x) - g(x) $$ and we see that in order to assess the limit as $x\to 0$, we don't need to know more about $h$ (it is bounded, so $x\cdot h(x)\to 0$), but we do need to know more about $g(x)$. Of course, it is easy to go back and check that $g(x) = -\frac8{3!} + x^2\cdot g_1(x)$ for some function $g_1$ that is bounded for $x\to 0$. Which is enough to conclude that the limit is indeed $\frac13$.

Arthur
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  • Honestly, I find the $O$ notation more intuitive than your functions $g,h$. – Miguel Nov 01 '20 at 15:24
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    @Miguel That may be. I still stand by my claim here, though: To a person who hasn't seen them before, arithmetic with an $O$ term is something new (and not entirely obvious) that one has to learn, while arithmetic with the $g$ and $h$ functions works exactly the way one is used to already from years of algebra in school. – Arthur Nov 01 '20 at 15:28
  • I explain $O(x^3)=k x^3$ for an unknown $k$. It seems easier than some arbitrary function, that also must be bounded. But of course it is a matter of what you see obvious. – Miguel Nov 01 '20 at 15:34
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    @Miguel I'm not saying the entire notion is easier to understand. I'm just claiming that concrete calculations are easier, because there aren't any new rules one must learn. – Arthur Nov 01 '20 at 15:36
  • We risk being moved to chat :) but $O(x^3)/x=O(x^2)$ does not seem a new rule at all. (Admittedly, $2 O(x^3)=O(x^3)$ requires some thought). But I think the problem is the same: to think of abstract, unknown functions. – Miguel Nov 01 '20 at 15:38
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    @Miguel I would say it is a new rule. Regardless, $\frac{O(x^3)}{2} = O(x^3)$ is clearly a new rule. As is $O(x^3) + x^4 = O(x^3)$. It also makes it rather difficult to do what I did at the end and go back to add more terms, without doing everything over again. Also, $g$ and $h$ aren't anywhere near as abstract as $O$ is. $O$ isn't even a function, even though it looks like one. – Arthur Nov 01 '20 at 15:39
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    @Miguel: $O(x^3)/x = O(x^2)$ is certainly a very intuitively appealing rule. But to use $O(-)$ notation correctly, one needs to learn that intuitively appealing rules aren’t always true — e.g. the same kind of intuition that makes $O(x^3)/x = O(x^2)$ “not new” can also mislead novices into thinking $O(x^2)-O(x^2) = 0$. – Peter LeFanu Lumsdaine Nov 01 '20 at 23:43
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In all the cases we should use remainder to proceed properly as follows

$$\frac{2 x \cos x- \sin 2x}{x^3}=\frac{2 x \left(1-\frac12 x^2+O(x^3)\right)- \left(2x-\frac16 (2x)^3+O(x^4)\right)}{x^3}=$$

$$=\frac{2x-x^3-2x+\frac43x^3+O(x^4)}{x^3}=\frac13+O(x) \to \frac13$$

without remainder we can easily get wrong with the solution.

user
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