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$$\lim _{x\rightarrow 0}{\frac {\cos \left( x \right) \sin \left( x \right) -x}{ \left( \sin \left( x \right) \right) ^{3}}}$$

I know that the real limit is $-2/3$ However, I've noticed that by approximating $\sin(x)$ as $x$ and $\cos(x)$ as $1-(x^2/2)$ I get the following:

$(1-(x^2/2))x - x)/ x^3 = (1- (x^2/2) -1 )(1/x^2) = -x^2/(2x^2) = -1/2 $

also if I only partially approximate x like: $(x\cos(x) - x)/(x^3)$ = $(\cos(x)-1)/x^2$ and then use L'hospital's rule to chisel this down I get: (L'hospital) = $-\sin x / 2x $ = (L'hospital) = $-\cos(x) / 2 = -1/2 $

Why does this conflict with doing L'hospital the whole way through without approximating $\sin(x)$ as $x$ ? Why is approximating $\sin(x)^3$ as $x^3$ wrong? Isn't this always approaching zero?

Asaf Karagila
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nofe
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  • @AsafKaragila thanks for the edit (i'm studying in B7 also) – nofe Feb 04 '12 at 10:56
  • But you study physics. Oy vey. :-) – Asaf Karagila Feb 04 '12 at 11:02
  • How do you know the "real" limit is $-2/3$? – Gerry Myerson Feb 04 '12 at 11:17
  • i did only l'hospital and that's what I got...and i confirmed with Maple – nofe Feb 04 '12 at 11:25
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    All your problems would be solved if you actually wrote the terms you discarded in the form of $o(x^\alpha)$. Besides having an actual equality between the two terms (instead of a poorly defined $\simeq$), you'd then notice that you get a term that looks like $o({1 \over x})$, and as $x \rightarrow 0$, this doesn't necessarily converge to zero. – Najib Idrissi Feb 04 '12 at 12:27

4 Answers4

27

As you know, $\sin x$ is only approximately equal to $x$, and $\cos x$ is only approximately $1 - \frac12 x^2$. How do you know when the approximation is good enough? One way is to keep track of how big the terms we've thrown away are. If you look at the Taylor series of $\sin$ and $\cos$, you find that $\sin x = x + O(x^3)$, and $\cos x = 1 - \frac12 x^2 + O(x^4)$, where $O(x^n)$ means something on the order of $x^n$ whose exact value we don't care about. So your limit is $$\begin{align} \lim_{x\to 0} \frac {\cos x \sin x - x}{\sin^3 x} &= \lim_{x\to 0} \frac{\big(1 - \frac12 x^2 + O(x^4)\big)\big(x + O(x^3)\big) - x}{\big(x + O(x^3)\big)^3} \\ &= \lim_{x\to 0} \frac{\big(x - \frac12 x^3 + O(x^3)\big) - x}{x^3 + O(x^5)} \\ &= \lim_{x\to 0} \frac{-\frac12 x^3 + O(x^3)}{x^3 + O(x^5)} \\ &= \lim_{x\to 0} -\frac12 + O(1) \end{align}$$ As you can see, one of the terms we ignored produces an error that doesn't go away as we approach $0$, so our solution is no good. If you trace backward to where the error came from, you'll find that it's the $O(x^3)$ term in the approximation of $\sin$. Then you would be wise to replace it with its true value, giving $\sin x = x - \frac16 x^3 + O(x^5)$, and evaluate the limit again. This time you should get the right answer, with an extra term that goes to zero as $x \to 0$.

6

You've lost a term involving $x^3$ by approximating $\cos x\sin x$ as $(1-(x^2/2))x$ instead of $(1-x^2/2)(x-(x^3/6))$.

Gerry Myerson
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  • i've done this over 4 times in different variations and each time I approximate $\sin(x)$ as $x$ I get a final limit value of $-(1/2)$ as opposed to $-(2/3)$ – nofe Feb 04 '12 at 11:28
  • i'm not sure I understand what you wrote...what do you mean I've lost a term? I made an algebraic error or I did something that isn't allowed in approximations (and i'd like to know what that specific transgression is) – nofe Feb 04 '12 at 11:31
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    Look at the two things on either side of the "instead of" in my answer. The first is what you wrote. The second is different. See the difference? That's the term you lost. Work out what happens when you use what I wrote instead of using what you wrote. – Gerry Myerson Feb 04 '12 at 11:52
  • ok after reading what everyone wrote here I see the mistake. Is it safe to assume then that when approximating terms I shouldn't "throw away" a "relevant term," like since in this particular problem I have $x^3$ in the denominator so when doing the taylor expansion of $sin(x)$ I should begin my $o(x^n)$ term after I have a value for the $x^3$ spot, namely $(x^3)/6$. – nofe Feb 04 '12 at 14:04
  • Yes, that's the point. – Gerry Myerson Feb 04 '12 at 22:53
  • @Rahul, thanks for the correction. – Gerry Myerson Feb 04 '12 at 22:56
  • @nofe: I would say that it's not so much 'you shouldn't throw away a relevant term' as that 'there are no hard and fast rules for knowing what's relevant'. It's not the $\sin^3 x$ in the denominator that's the problem here as much as the fact that the numerator is essentially cubic in $x$. The best way to do it isn't by throwing away terms at all, but by including your $O()$ terms in your calculation so you know what your error bounds are. – Steven Stadnicki Feb 26 '12 at 17:40
2

I'm not going about proving it here (the people above have already provided good solutions), I'm just explaining why you can't always apply limits directly inside a limit:

The point is that, sin x and cos x tend to their approximations at different rates. So, their difference will not tend to the difference of their limits.

As a general rule of thumb, you can do:

  1. $\lim ab=\lim a \lim b$,

  2. $\lim \frac{a}{b}=\frac{\lim a}{\lim b}$,

  3. and $\lim a+b=\lim a+\lim b$,

  4. but not $\lim \frac{a+b}{c}=\frac{\lim a+\lim b}{\lim c}$

even though the last limit appears to be a combination of the second two. The flaw here is that we can only apply $\lim \frac{a}{b}=\frac{\lim a}{\lim b}$, when $\lim b$ is defined and not zero. So, $\lim \frac{a+b}{c}=\frac{\lim a+\lim b}{\lim c}$ only works when $\lim c=0$, which is almost always not the case when you apply this.

Manishearth
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2

You need to relate this issue to the "order of approximation" of a Taylor series. You can show that if $P_n^a(x)$ is the Taylor Polynomial of order $n$ around $x=a$ of $F(x)$ then

$$\lim_{x\to a}\frac{F(x)-P_{n}^a}{(x-a)^n}=0$$

Just to make things clear $$P_n^a(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k$$

In $\mathcal{o}$ notation

$$f(x) = P_n^a(x)+o((x-a)^n)$$

Let's put this into practice:

We can readily check that

$$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1 - x}}{x} = 0$$

But what if we take

$$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1 - x}}{{{x^2}}}$$

Now the limit is not zero ($=1/2$). So what do we need to make things "right"? Adding the next term in the Taylor series does the work. So we have:

$$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1 - x - \frac{{{x^2}}}{2}}}{{{x^2}}} = 0$$

Similarily for the sine, when dealing with higher order powers of $x$, it doesn't suffice to put only $\sin x \sim x$. Although since for the sine the even powers are zero, some formulas do work, like

$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x}}{{{x^2}}} = 0$$

But for $x^3$ we need the next approximation:

$$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x - \frac{{{x^3}}}{3!}}}{{{x^3}}} = 0$$

Thus you see $\sin(x) = x + o(x^3) $ is not correct, although $\sin(x) = x + o(x^2) $ and $\sin(x) = x + o(x)$ are.

IMPORTANT: Do not confuse Big-O with Little-O notation.

Pedro
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