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I see some answers in this site proving many limits by Taylor, for example

$$\lim_{x \to 0}\dfrac{\sin x-x}{x^3}=-\dfrac16$$

comes to my mind. However, my teacher didn't teach me this technique, also the popular calculus/analysis books I have do not even mention this technique. When people use Taylor to compute this limit, they often use big-$O$ operations, or omitting the terms having high order. However, what are the big-$O$ techniques? How does they work? Why is this rigorous? And how to choose where the omitting occurs? Should I use Taylor "expansions" (with Lagrange remainder) or Taylor "series"? I totally have no idea. Need your help.

M. Winter
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Eric
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  • Use Taylor series when you want to be exact. Use Taylor polynomial with Landau rest term when you want to be exact enough in a rigorous sense. – M. Winter Oct 27 '17 at 14:17

3 Answers3

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Formally, a Taylor expansion up to the $k$-th term means that

$$f(x+h)=f(x)+hf'(x)+\cdots+\frac 1{k!}h^k f^{(k)}(x)+\color{red}{o(h^k)}$$

where the little $o$-term is a placeholder for a function $g(h)$ that i not further specified except for its property that

$$\lim_{h\to 0}\frac{g(h)}{h^k}=0$$

Because this fraction vanishes for small $h$, it vanishes in limits as the one in your example. This is useful in situations where we have an approximation for a certain function which is not perfect, but we want to express that we know this and that we can say how good/bad it approximates.

Example. Choose $f(x)=\sin(x)$ and observe

$$\sin(x)=\sin(0)+x\cos(0)-\frac{x^2}2\sin(0)-\frac{x^3}6\cos(0)+o(x^3)=x-\frac{x^3}6+o(x^3).$$

Now consider your example:

\begin{align} \frac{\sin(x)-x}{x^3} &=\frac{x-\frac{x^3}6+o(x^3)-x}{x^3}\\ &=\frac{-\frac{x^3}6+o(x^3)}{x^3}\\ &=-\frac16+\frac{o(x^3)}{x^3}\\ \end{align}

and we know that $o(x^3)/x^3\to0$ for $x\to 0$. So the limit is $-1/6$. See that we do not actually need to know what the function $o(x)$ stands for, as we only need that it vanishes in certain limits. This is the power of the Landau $\mathcal O$-notation (I could have used $\mathcal O(x^{4})$ instead, but I personally find $o(x^3)$ more appealing here). Of course the really powerful result here is Taylor's theorem which provides you with this $o(x^3)$ term in the first place.

M. Winter
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  • Thanks! One little question, in your first paragraph, is the red part(i.e., $g(h)$) be produced by Taylor expansion(namely, $f(x+h)=f(x)+hf'(x)+\cdots+\frac 1{k!}h^k f^{(k)}(x)+\frac{f^{(k+1)(\xi)}}{(k+1)!}h^{k+1}$, where $\xi$ between $x$ and $h$), or Taylor series? If this it the former, wouldn't the $\xi$ changes whenever $h$ changes? – Eric Oct 27 '17 at 16:19
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    @Eric If I remember correctly, then I know this first formula of my post as theorem of Taylor, but not sure. It gives a Taylor expansion up to the $k$-th term and information on the rest term. The Lagrange rest term can be considered a separate result which concretizes the form of the rest term $o(h^k)$. Taylor series are a result on the convergence of sequences of Taylor expansions and I think are a bit farther away from what I have talked about here. – M. Winter Oct 27 '17 at 16:24
  • I see. Seems that I ignored the so-called "Taylor theorem". This isn't mentioned in many of my textbooks... Let me google it. – Eric Oct 27 '17 at 16:26
  • Oh! I just read the another answer below. (I read the answers followed the order.) – Eric Oct 27 '17 at 16:29
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    @Eric I found it as Taylor's theorem and it essentially talks about the Taylor expansion. I think every well-known and important construct of math started with a theorem at some time ;) – M. Winter Oct 27 '17 at 16:30
  • Thanks. Very helpful. Both answers are the best answers in my mind! You are a good teacher! :) – Eric Oct 27 '17 at 16:58
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    @Eric I am glad that this helped you and thank you for your compliment :). I do not want to rush you but keep in mind to accept an answer. If you cannot decide which, then toss a coin. I looked at your question history and it seems you have a lot of questions with answers where none was accepted. Note that accepting answers is an important part of the StackExchange model as you can read here. It completes the post and allows other users to refer it. And on top of it you get +2 reputation each time ;). – M. Winter Oct 27 '17 at 20:56
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    @Eric: This particular theorem of Taylor may not be famous enough to be included in every calculus / analysis text, but it is the simplest of all the various versions of Taylor's theorem and requires minimal assumptions for the function $f$. You may wish to have a look at its proof in this answer: https://math.stackexchange.com/a/1809307/72031 – Paramanand Singh Oct 29 '17 at 00:59
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That is quite a few questions.

$O$-notation: A function $f(x) = O(g(x))$ as $x \to x_0$ if there exist a neighbourhood $B(x_0,\epsilon)$ of $x_0$ and $M>0$ such that $|f(x)| \leq M |g(x)|$, for $x \in B(x_0,\epsilon)$.

Connection to taylors formula with remainder: Using taylors formula with remainder we have $$ f(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k + \frac{f^{(n)}(z)}{n!}(x-x_0)^n, $$ where $z$ lies between $x$ and $x_0$. If $f^{(n)}$ is continuous near $x_0$ we have that $|f^{(n)}(z)| \leq \max_{z \in [x_0- \epsilon, x_0+\epsilon]}|f^{(n)}(z) |=M_1$. Setting $M=M_1/n!$ we get that:

$$ |f(x) - \sum_{k=0}^{n-1} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k| \leq M|(x-x_0)^n| $$ near $x_0$. Therefore $f(x)-\sum_{k=0}^{n-1} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k = O(x-x_0)^n$ as $x\to x_0$.

A little calculus with $O$-notation: Assume $f(x) = O((x-x_0)^n)$ as $x\to x_0$ and $0<k \leq n $ then $f(x)/(x-x_0)^k = O((x-x_0)^{n-k})$. This follows easily from the fact that in $B(x_0,\epsilon)$ we have that: $$ |\frac{f(x)}{(x-x_0)^k}| \leq \frac{M|(x-x_0)^n|}{|(x-x_0)^k|} = M|(x-x_0)^{n-k}|. $$

Now for your problem: We know from taylor (and the above connection with $O$-notation) that $\sin(x) = x-\frac16 x^3 +O(x^4)$ as $x\to 0$. Now we have that $\frac{\sin(x) -x}{x^3} -(- \frac16) = O(x)$. That is $|\frac{\sin(x) -x}{x^3} - (-\frac16)| \leq M |x| $ as $x \in B(0,\epsilon)$. Thus $\lim_{x\to 0}\frac{\sin(x) -x}{x^3}=-\frac16$.

(Btw: $f(x)=g(x)+O(h(x)$ means that $f(x)-g(x) = O(h(x))$)

Martin
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  • Thanks! Very, very, clear! Pretty valuable answer. Both answers help me a lot. Can I ask what textbooks have mentioned these things? – Eric Oct 27 '17 at 16:56
  • I learned real analysis from T.M.Apostol's book "Mathematical Analysis". But he doesn't use $O$-notation much as far as I recall. – Martin Oct 27 '17 at 19:16
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Hint: $\sin(x)=x-x^3/6+ O(x^4)$ This implies that ${{\sin(x)-x}\over x^3}=-1/6+O(x)$.

https://en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions

Tsemo Aristide
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    I don't think this answers the question. –  Oct 27 '17 at 13:43
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    The OP doesn't seem too familiar with Big-$O$ notations and the meaning of $O(x)$ for a Taylor series. You should expand your answer. – Zubzub Oct 27 '17 at 13:44