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We are given

$$ \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $$

We can write

$$ L = \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} = \lim _{x \to 0} \frac {x (cos x - \sin x/x)} {x^2 \sin x} $$

It can be written as

$$ L = \lim _{x \to 0} \frac { \cos x - \sin x/x} {x^2 \sin x/x} $$

We know

$$ \lim _{x \to 0} \frac {\sin x} {x} = 1 $$

The original limit can be written as

$$ L = \lim _{x \to 0} \frac { \cos x - 1} {x^2} $$

Now using $ ( \cos x ) $ expansion formula, the answer will be $ L = -\frac12$ , but the answer is $ L = -\frac13$ , which is not matching.

I want to know what I did wrong in this method. I know the answer using other methods, but I want to know what went wrong with this one.

Rodrigo de Azevedo
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Ga Me
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    I don't think that you can replace $\sin(x)/x$ with 1 like that. That is, I don't think you can say $\lim_{x\to 0} \frac{\cos(x)-\sin(x)/x}{x^2\sin(x)/x} = \lim_{x\to 0}\frac{\cos(x)-1}{x^2}$ – gist076923 Dec 20 '22 at 14:34
  • ^^ $\to \lim_{x \to 0}{x^2\frac{\sin{x}}{x}} ≠ \lim_{x \to 0}{\frac{\sin{x}}{x}}$. See, section 4 under https://tutorial.math.lamar.edu/classes/calci/limitsproperties.aspx. Additionally, here is a possible duplicate: https://math.stackexchange.com/questions/3772486/find-limits-of-lim-x-to-0-frac-x-cos-x-sin-x-x2-sin-x-without?rq=1 – Dstarred Dec 20 '22 at 14:37
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    Does this answer your question? Find limits of $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor Expansion. – D S Dec 20 '22 at 15:04
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    The main problem is that when $x\to 0$, it's true that $\frac{\sin(x)}{x}$ tends to $1$ but $\cos(x)$ also goes to 1 simulatenously. Thus we have the case that $\lim_{x\to 0}\frac{\cos(x)}{x^{2}\frac{\sin(x)}{x}}$ not existing as well as $-\frac{1}{x}$ not existing but you are adding them up. This is a fallacy . In particular if $\lim f(x)$ and $\lim g(x)$ both exist then you can conclude that $\lim f(x)+g(x)=\lim f(x)+\lim g(x)$ and same way for products . i.e . $\lim f(x)g(x)=\lim f(x)\cdot\lim g(x)$ . So upto your third line, it's correct. But after that everything goes wrong. – Mr.Gandalf Sauron Dec 20 '22 at 15:34
  • To illustrate what I am saying, $\lim_{x\to 0}(\frac{1}{x}-\frac{1}{x})=\lim_{x\to 0} 0=0$ . But $\lim_{x\to 0}\frac{1}{x}$ does not exist. However if someone asks me $\lim_{x\to 0}\frac{\sin(x)-\tan(x)}{x}$ . The I can say that as $\lim_{x\to 0}\frac{\sin(x)}{x}=\lim_{x\to 0} \frac{\tan(x)}{x}=1$ we have the answer is $1-1= 0 $ or even $1-\lim_{x\to 0}\frac{\tan(x)}{x}=0$ . The difference is that both the limits exist independently . – Mr.Gandalf Sauron Dec 20 '22 at 15:40
  • @DS this is not a duplicate because the question here is not "what is the limit?" but "where is my mistake?". – Anne Bauval Dec 20 '22 at 15:42
  • @UnexpectedConfusion this is not a duplicate, as just explained in my previous comment. – Anne Bauval Dec 20 '22 at 15:45
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    See https://math.stackexchange.com/a/1783818/72031 – Paramanand Singh Dec 21 '22 at 00:32

4 Answers4

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Your mistake was to replace $\frac {\sin x}x$ by its limit and then take again the limit of the resulting expression.

Generally, as your example shows, $$\lim_{x\to a}f(x,g(x))\ne \lim_{x\to a}f(x,\lim_{x\to a}g(x)).$$ A simpler counterexample is: $$\lim_{x\to0}\frac xx=1\ne0= \lim_{x\to0}\frac{\lim_{x\to0}x}x.$$

Anne Bauval
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To check where it went wrong , let us use Series Expansion :

$ L = \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $
$ L = \lim _{x \to 0} \frac { x ( 1-x^2/2!+x^4/4!-x^6/6! \cdots ) - ( x-x^3/3!+x^5/5!-x^7/7! \cdots ) } {x^2 ( x-x^3/3!+x^5/5!-x^7/7! \cdots ) } $

$ L = \lim _{x \to 0} \frac { ( x-x^3/2!+x^5/4!-x^7/6! \cdots ) - ( x-x^3/3!+x^5/5!-x^7/7! \cdots ) } { ( x^3-x^5/3!+x^7/5!-x^9/7! \cdots ) } $

$ L = \lim _{x \to 0} \frac { -x^3/3 + (x^5/4!-x^7/6! \cdots ) - (x^5/5!-x^7/7! \cdots ) } { ( x^3 + (-x^5/3!+x^7/5!-x^9/7! \cdots) ) } $

$ L = \lim _{x \to 0} \frac { -1/3 + (x^2/4!-x^4/6! \cdots ) - (x^2/5!-x^4/7! \cdots ) } { ( 1 + (-x^2/3!+x^4/5!-x^6/7! \cdots ) ) } $ [[ Cancelling $x^3$ through out ]]

$ L = \lim _{x \to 0} \frac { -1/3 + 0 - 0 } { ( 1 + 0 ) } $ [[ Setting $x=0$ through out ]]

$ L = -1/3 $ & there is no Doubt about this Answer.

Now we get a clue or hint about what went wrong.

The given limit is $ ( \infty - \infty ) $ where we can not take the Individual limits.

Numerator is in the format $ (x+A_1x^3+A_2x^5 \cdots ) - (x+B_1x^3+B_2x^5 \cdots ) $

Denominator is in the format $ (x^3+C_1x^5+C_2x^7 \cdots ) $

Numerator will be cancelling the $ (x) - (x) $ terms.
Denominator & Numerator will then have Common $ (x^3) $ which will be cancelling throughout.

We want both $ A_1 = -1/2 $ & $ B_1 = -1/6 $ to get the Exact limit $ L = A_1 - B_1 = (-1/2) - (-1/6) = (-1/3) $ which is Correct.
Whereas by taking Individual limit earlier , we are keeping $ A_1 = -1/2 $ while losing $ B_1 = -1/6 $ , hence getting the limit $ L = A_1 = -1/2 $ which is wrong.

Developing that Core Point with Extreme Detail :

We have $ L = \frac{ (x+A_1x^3+A_2x^5 \cdots ) - (x+B_1x^3+B_2x^5 \cdots ) }{ (x^3+C_1x^5+C_2x^7 \cdots ) } $

In the Denominator , we take $x^3=x \cdot x^2$ & leave the $x^2$ in the Denominator , to use the $x$ on the Numerator $A$ & $B$ terms to get :
$ L = \frac{ (x+A_1x^3+A_2x^5 \cdots )/x - (x+B_1x^3+B_2x^5 \cdots )/x }{ (x^2+C_1x^4+C_2x^6 \cdots ) } $

We have $ L = \frac{ \cos x - (1+B_1x^2+B_2x^4 \cdots ) }{ (x^2+C_1x^4+C_2x^6 \cdots ) } $

It is Correct till here.

The trouble starts now.
We are now using $ ( \sin x ) / ( x ) = 1 $ , thereby losing $B_1$ , yet keeping $A_1$ in the $ ( \cos x ) $ term. That is WRONG ! We have to Consistently keep both terms having the Same Power !

When we take over-all limit while keeping both $A$ term & $B$ term having the $x^3$ Power , we get $L=-1/3$ , which is Correct.

Prem
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You should not be taking a limit, then taking it again for what you get after having taken the first limit. The correct way of evaluating the limit would be to split the fraction up at the point of $\frac{\cos{x}-\sin{x}/x}{\sin{x}/x},$ and to proceed from there, eventually applying the Taylor expansion, or just to use l'Hopital's rule.

IChoi
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The thing is you can take a partial limit when it only involves products (or quotients). Here there is an additive $\cos(x)$ term so you cannot do that.

Remember that $\cos(x) = 1 - \frac{x^2}{2} + o(x^2)$ and $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + o(x^2)$ therefore $\cos x - \frac{\sin x}{x} = -\frac{x^2}{3} + o(x^2)$ which means that $ \frac { \cos x - \sin x/x} {x^2 }\to -\frac{1}{3}. $

Therefore $$\lim_{x \to 0} \frac { \cos x - \sin x/x} {x^2 \sin x/x} = \lim_{x \to 0} \frac { -\frac{1}{3} } {\sin x/x} = -\frac{1}{3} $$

Lelouch
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