<s>[INST] What is wrong with the calculation $ \lim\limits_{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x} $? 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.
 [/INST] 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.
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