Why does $(\sin x)^2=x^2$ and $\sin x=x$ in these contexts?

Question

Contexts (it must also be noted that as $\delta t$ tends to zero, $\delta \theta$ also tends to zero):

First context $$ \lim_{\delta t \to 0} \frac{-2v\sin(\delta\theta/2)^2}{\delta t} = \lim_{\delta t \to 0} \frac{-2v(\delta\theta/2)^2}{\delta t}\quad\text{[since $\sin(x) \to x$ as $x \to 0$]} $$ Second context $$ \lim_{\delta t \to 0} \frac{v\sin(\delta\theta) - 0}{\delta t} = \lim_{\delta t \to 0} \frac{v\sin(\delta\theta)}{\delta t} = \lim_{\delta t \to 0} \frac{v\delta\theta}{\delta t} = v\omega = r\omega^2\quad\text{[since $\sin(x) \to x$ as $x \to 0$]} $$ The only reason that I am given is that as $x$ tends to zero, $\sin x$ tends to $x$, which for one does not only not make sense, but also doesn't help to clarify why $(\sin x)^2=x^2$ in the first context. So, I was wondering if someone could either provide me with the correct explanations or explain the ones provided.

Answer 1

As @NoahSchweber pointed out, strictly speaking, $\sin(x) \to x$ is nonsense; what it means is that $\sin(x)/x \to 1$. This indicates that, with care (see @Bernard's post for details), you may replace $\sin(x)$ by $x$ in limits. Assuming that $\Delta\theta \to 0$ as $\Delta t \to 0$ (EDIT: as I see that you said is the case), you have, for example, $$ \lim_{\Delta t \to 0} \frac{-2v\sin^2(\Delta\theta/2)}{\Delta t} = \lim_{\Delta t \to 0} \Bigl(\frac{\sin(\Delta\theta/2)}{\Delta\theta/2}\Bigr)^2\cdot\frac{-2v(\Delta\theta/2)^2}{\Delta t} = \lim_{\Delta t \to 0} \frac{-2v(\Delta\theta/2)^2}{\Delta t}, $$ as long as the last limit exists, which it need not without an additional hypothesis (namely, the differentiability of $(\Delta\theta)^2$ as a function of $\Delta t$ at $\Delta t = 0$).

Answer 2

The rigourous mathematical notion is equivalence of functions in a neighbourhood of $0$. Roughly speaking, two functions $f$ and $g$ are equivalent near $0$ if $\;\lim_{x\to 0}f(x)/g(x)=1$.

Now equivalence of functions is compatible with multiplication and division (but not with addition and substraction), and a function has limit $\ell\ne 0$ when $x$ tends to $0$ if and only if it is equivalent to $\ell$ near $0$.

Answer 3

I have pointed out in another answer here that replacing $A$ by $B$ when $A \neq B$ in any sort of mathematical derivation is wrong. The idea of replacing $\sin x$ by $x$ when we are dealing with a limit $x \to 0$ is plain wrong. What we can do is if we see an expression like $\lim_{x \to 0}\dfrac{\sin x}{x}$ then we can replace it by $1$ because these two things are equal.

But consider the following: \begin{align} L &= \lim_{x \to 0}\frac{\log(1 + x)}{\sin x}\tag{1}\\ &= \lim_{x \to 0}\dfrac{\log(1 + x)}{x\cdot\dfrac{\sin x}{x}}\tag{2}\\ &= \lim_{x \to 0}\dfrac{\log(1 + x)}{x}\cdot\dfrac{1}{\lim_{x \to 0}\dfrac{\sin x}{x}}\tag{3}\\ &= \lim_{x \to 0}\dfrac{\log(1 + x)}{x}\cdot\dfrac{1}{1}\tag{4}\\ &= \lim_{x \to 0}\dfrac{\log(1 + x)}{x}\tag{5} \end{align} If we just look at the steps $(1), (5)$ without looking at the intermediate steps $(2)-(4)$ then it appears that $\sin x$ in denominator has been replaced by $x$. In reality $\sin x$ is never replaced by $x$ because they are not equal unless $x = 0$ (which is not the case here) and wherever you see such replacement think of it as some intermediate steps being removed because the author thinks that these intermediate steps are obvious enough for the reader to create in his imagination.

Unless one is somewhat experienced in the art of calculus it is better not to miss the steps $(2)-(4)$ in any derivation. Also in my opinion it is not a good idea to load students with rules regarding justification of such replacements in certain contexts. A student is better off when he sticks to the philosophical principle that $A$ is $A$ and it cannot be replaced by $B$ unless $A = B$. When enough maturity is attained the student will not ask why such replacements are OK, but he will imagine the intermediate steps in his mind and continue as if this does not need any explanation.