Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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?

share|cite|improve this question
@AsafKaragila thanks for the edit (i'm studying in B7 also) – nofe Feb 4 '12 at 10:56
But you study physics. Oy vey. :-) – Asaf Karagila Feb 4 '12 at 11:02
How do you know the "real" limit is $-2/3$? – Gerry Myerson Feb 4 '12 at 11:17
i did only l'hospital and that's what I got...and i confirmed with Maple – nofe Feb 4 '12 at 11:25
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 4 '12 at 12:27
up vote 23 down vote accepted

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 $x \to 0$.

share|cite|improve this answer

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

share|cite|improve this answer
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 4 '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 4 '12 at 11:31
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 4 '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 4 '12 at 14:04
Yes, that's the point. – Gerry Myerson Feb 4 '12 at 22:53

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.

share|cite|improve this answer
thanks, that was helpful – nofe Feb 4 '12 at 15:12

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.