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The following is a problem from Chapter 20 of Spivak's Calculus.

  1. Find the following

(e) $\lim\limits_{x\to 0} \left (\frac{1}{\sin^2{(x)}}-\frac{1}{\sin{(x^2)}}\right )$

The aim of this problem is to use Taylor Polynomials and the Taylor Theorem to avoid using L'Hôpital's Rule.

The solution manual simply says

$$\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}\tag{1}$$

$$=\lim\limits_{x\to 0} \frac{\left ( x^2-\frac{x^6}{3!}+... \right )-\left ( x-\frac{x^3}{3!}+... \right )^2}{\left ( x^2-\frac{x^6}{3!}+... \right ) \left ( 1-\frac{x^3}{3!}+... \right )^2}$$

$$=\lim\limits_{x\to 0}\frac{\frac{2x^4}{3!}}{x^2}\tag{2}$$

$$=0$$

I'd like to understand the steps that lead to $(2)$.

Below is my attempt. Note that after I went through the steps below a few times, I finally remembered I could check the limit in Maple. Maple gives the same result that I obtain below, $\frac{1}{3}$.

In light of this, my question is: is the Spivak solution manual really incorrect? It would seem so, though it has happened in the past that I have thought I identified a mistake in the solution manual only to realize I had been solving the wrong problem in some way. If the solution manual is indeed incorrect, is the solution below correct?

By Taylor's Theorem we can write

$$\sin{(x)}=P_{2n+1,0,\sin}(x)+R_{2n+1,0,\sin}(x)$$

$$=\sum\limits_{i=0}^n (-1)^i \frac{x^{2i+1}}{(2i+1)!}+R_{2n+1,0,\sin}(x)$$

and it can be shown that

$$P_{n,0,\sin^2{(x)}}=[P_{n,0,\sin}\cdot P_{n,0,\sin}]_n$$

where $[P]_n$ denotes the truncation of the polynomial $P$ to degree $n$, ie the sum of all terms of $P$ of degree $\leq n$.

In addition,

$$P_{n,0,\sin{(x^2)}}(x)=\left[\sum\limits_{i=0}^n (-1)^{i} \frac{x^{i+1}}{i!}\right ]_n=\left [ x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-... \right ]_n$$

Let's consider fifth order Taylor polynomials of the functions $\sin^2{(x)}$ and $\sin{(x^2)}$, and let's sub into $(1)$.

We have

$$\sin{(x^2)}=x^2+R_{5,0,\sin{(x^2)}}(x)$$

$$\sin^2{(x)}=x^2-\frac{x^4}{3}+R_{5,0,\sin^2}(x)$$

If we sub into $(1)$, we get the computations below. They look long because of the notation I am using, but it is simple multiplication essentially.

$$\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}\tag{1}$$

$$=\lim\limits_{x\to 0} \frac{x^2+R_{5,0,\sin{(x^2)}}(x)-(x^2-\frac{x^4}{3}+R_{5,0,\sin^2}(x))}{(x^2-\frac{x^4}{3}+R_{5,0,\sin^2}(x))(x^2+R_{5,0,\sin{(x^2)}}(x))}$$

$$=\lim\limits_{x\to 0} \frac{R_{5,0,\sin{(x^2)}}(x)+\frac{x^4}{3}-R_{5,0,\sin^2}(x)}{x^4+x^2 R_{5,0,\sin{(x^2)}}(x)-\frac{x^6}{3}-\frac{x^4 R_{5,0,\sin{(x^2)}}(x)}{3}+x^2R_{5,0,\sin^2}(x)+R_{5,0,\sin^2}(x)R_{5,0,\sin{(x^2)}}(x)}$$

$$=\lim\limits_{x\to 0} \frac{\frac{1}{3}+\frac{R_{5,0,\sin{(x^2)}}(x)}{x^4}-\frac{R_{5,0,\sin^2}(x)}{x^4}}{1+\frac{R_{5,0,\sin{(x^2)}}(x)}{x^2}-\frac{x^2}{3}-\frac{R_{5,0,\sin{(x^2)}}(x)}{3}+\frac{R_{5,0,\sin^2}(x)}{x^2}+\frac{R_{5,0,\sin^2}(x)}{x^4}+\frac{R_{5,0,\sin^2}(x)}{x^2}\frac{R_{5,0,\sin{(x^2)}}(x)}{x^2}}$$

$$=\frac{1}{3}$$

where I used the fact that each remainder is a polynomial of degree 6.

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  • $\begingroup$ The given answer is saying there is a typo that was propagated. There is nothing else wrong. $\endgroup$ Commented Oct 2, 2022 at 7:27
  • $\begingroup$ Your method is more consistent with what Spivak's book teaches (and it is correct); in particular, referencing all of the work from Problem 9. I would note, though, that you needed to only go out to the 4th degree polynomial (rather than 5th) $\endgroup$
    – S.C.
    Commented Feb 9 at 5:31

2 Answers 2

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I would use $${\sin(x^2)-\sin ^2x\over x^4} \cdot {x^2\over \sin(x^2)}\cdot {x^2\over \sin^2x}$$ The last two factors tend to $1.$ Next $$\displaylines{\sin(x^2)-\sin^2x=\sin(x^2)+{1\over 2}[\cos2x-1]\\ = x^2+o(x^4)+{1\over 2} \left [-{(2x)^2\over 2!}+{(2x)^4\over 4!}+o(x^4)\right ]\\={1\over 3}x^4+o(x^4)}$$ Hence $$\lim_{x\to 0}{\sin(x^2)-\sin ^2x\over x^4}={1\over 3}$$

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  • $\begingroup$ We can continue in the same vein by rearranging the limit to $\frac{1}{\sin\left(x\right)^{2}}-\frac{1}{\sin\left(x^{2}\right)}=\frac{b\left(x\right)\left(1+a\left(x\right)\right)-x^{2}b\left(x^{2}\right)}{a\left(x\right)^{2}a\left(x^{2}\right)}$ where $a\left(x\right)=\frac{\sin\left(x\right)}{x}\to1$ and $b\left(x\right)=\frac{x-\sin\left(x\right)}{x^{3}}\to\frac16$ are standard limits, avoiding any calculus methods entirely but defeating the purpose of the exercise. $\endgroup$
    – Jam
    Commented Oct 2, 2022 at 10:46
  • $\begingroup$ @Jam You are right. That's why I saved some elements of the Maclaurin expansion. $\endgroup$ Commented Oct 2, 2022 at 13:43
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Yes, or simply: $$\begin{align}\lim\limits_{x\to 0} \frac{\sin{(x^2)}-\sin^2{(x)}}{\sin{(x^2)}\sin^2{(x)}}&=\lim\limits_{x\to 0} \frac{\left ( x^2-\frac{x^6}{3!}+... \right )-\left ( x-\frac{x^3}{3!}+... \right )^2}{\left ( x^2-\frac{x^6}{3!}+... \right ) \left (\color{red}x-\frac{x^3}{3!}+... \right )^2}\\&=\lim\limits_{x\to 0} \frac{\left ( x^2+o(x^4)\right )-\left ( x^2-2x\frac{x^3}{3!}+o(x^4)\right )}{\left ( x^2+... \right ) \left (x^\color{red}2+... \right )}\\&=\lim\limits_{x\to 0}\frac{\frac{2x^4}{3!}}{x^\color{red}4}\\ &=\frac13.\end{align}$$

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  • $\begingroup$ But how do you go from the second expression to the third expression (ie from the first to the second line)? For example, how do you obtain $x^4$ in the denominator? $\endgroup$
    – xoux
    Commented Oct 2, 2022 at 7:29

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