$\lim_{x\to0} (\sin x-x)/x^3$ why can't the fraction be split first like $x/x^3 - \sin x/x^3$

Question

$\lim_{x\to0} (\sin x-x)/x^3$ why can't the fraction be split first like $x/x^3 - \sin x/x^3$ and then later on be written as $1/x^2 -\sin x/x \cdot x^2$ and $\sin x/x =1$ so $1/x^2-1/x^2=0$

Answer 1

Well, it simply doesn't work that way. You can write $$(\sin x-x)/x^3=\frac{(\sin x)/x}{x^2}-\frac1{x^2}$$if you like, but that doesn't help find the original limit, since the limits of the two terms on the right do not exist. Your argument seems to be saying that $$\lim_{x\to0}\frac{\sin x}{x}\frac1{x^2}=\frac1{x^2},$$but that's not true. (Makes no sense; if that limit did exist it would be a single number, not a function of $x$...)

Good of you to add the "explanation" $\sin(x)/x=1$. That's not true, in fact it's easy to see that $|\sin(x)/x|<1$ for every $x\ne0$.

Answer 2

$$\frac{\sin x-x}{x^3} = \frac{\frac{\sin x}x-1}{x^2}$$

You can't then simply say

$$\lim_{x\to0} \frac{\frac{\sin x}x-1}{x^2} \stackrel{?}{=} \lim_{x\to0} \frac {1-1}{x^2} = \lim_{x\to0} \frac 0{x^2} = \lim_{x\to0} 0= 0$$

Although the numerator goes to $0$, so does the denominator. The numerator does not get to be zero without the denominator being zero. You can't pick and choose, letting $\frac{\sin x}x$ become $1$ first before $x^2$ becomes $0$. In fact, you don't know which one goes to $0$ faster, the numerator or the denominator, so you can't tell if it goes to $0$ or $\pm\infty$ or anything inbetween.

Answer 3

Well, what you can do is split it up like

$$\lim_{x\to0}\frac{\sin x-x}{x^3}=\lim_{x\to0}\left(\frac{\sin x}{x^3}-\frac{1}{x^2}\right).$$

However you cannot distribute the limit, i.e.

$$\lim_{x\to0}\left(\frac{\sin x}{x^3}-\frac{1}{x^2}\right)\neq\lim_{x\to0}\frac{\sin x}{x^3}-\lim_{x\to0}\frac{1}{x^2}.$$

The reason for this is because to do this, you need both the limits on the right hand side to exist, and clearly they do not (for one, $\frac{1}{x^2}\to\infty$ as $x\to0$). This is a general property of limits: you can only distribute the limit if the limits exist (and are finite of course), otherwise it might not hold. Take for example the limit

$$\lim_{x\to\infty}(x-x)=\lim_{x\to\infty}0=0.$$

If we would have instead distributed the limit first thing we did, then we would end up with something undefined as $\infty-\infty$, which is clearly nonsense. While this example is extremely contrived and trivial, it still shows the point, and a lot of the situations do end up occuring, which is why you have to be careful.

Answer 4

You seem to be doing the following: $$ \lim\limits_{x\rightarrow 0}\frac{\sin(x)-x}{x^3}= \lim\limits_{x\rightarrow 0}\left(\frac{\frac{\sin(x)}{x}}{x^2}-\frac{1}{x^2}\right)\stackrel{\text{$\ast$}}{=} \lim\limits_{x\rightarrow 0}\left(\frac{\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{x}}{x^2}-\frac{1}{x^2}\right)= \lim\limits_{x\rightarrow 0}\left(\frac{1}{x^2}-\frac{1}{x^2}\right)= \lim\limits_{x\rightarrow 0}0= 0 $$

The problem is in the step marked by the $\ast$, all other steps are fine. Any step in math needs to be justified by a rule which is applied appropriately in the context of the expression. There is no rule for doing what is happening in step $\ast$. If even a single step cannot be justified, it might give you a faulty equality (which happens here) and all remaining steps, although correct when viewed in isolation, will in general produce unequal terms.

What you could do instead is to use L'Hôpital's rule. Alternatively you could expand $\sin(x)-x=-\frac{x^3}{6}+\sum\limits_{n=2}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}$ into a convergent power series and compute with that - this however might be beyond your level, as you need to interchange a limit and a series, which can be done here (theory of analytic functions which implies uniform convergence near $0$ for the function in question), but in general is wrong.

Your issues are in spirit similar to this beginner struggling with limits or that one.