0

I wrote out a proof using the definition of the derivative for the function $f(x) = \sin(\theta)$ to show that $f'(x) = \cos(\theta)$, in order to use L'H's rule to show that the limit, $x \to 0$, of $\frac{\sin(x)}{x}$ equals 1, but as one professor in my department warned me: this uses circular reasoning. There is a factor of $\frac{\sin(h)}{h}$ when using the definition

$$ f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h},$$

and we must show that

$$\lim_{h \to 0} \frac{\sin(h)}{h} =1,$$

to continue the proof.

So, how can we still justify using L'H's rule for $\lim_{x \to 0} \frac{\sin(x)}{x}$?

We're using the result to prove the result, which is absurd.

Thanks,

rtybase
  • 16,907
D.Hutchinson
  • 1
  • https://math.stackexchange.com/questions/707469/how-to-prove-that-d-sinx-dx-cosx-without-circular-logic-such-as-lh%C3%B4pi?rq=1 – John Doe Sep 28 '17 at 18:12
  • Whether it's circular reasoning depends on how you define the sine function. If you define it as $\sin x=\sum_{k\ge0}\frac{(-1)^kx^{2k+1}}{(2k+1)!}$, then there's no circularity at all, because the derivative is $\cos x$ by term by term differentiation and by definition of cosine. – egreg Sep 28 '17 at 20:20
  • It appear that the charm of L'Hospital's Rule will never diminish. Please see this answer. – Paramanand Singh Sep 29 '17 at 02:48

1 Answers1

0

there are few ways to show that $sin(x)= x$ when x is a small angle.

the 2 ways that i founds the most clear are those:

way 1:

small angle $$$$in this picture(taken from wiki:source of picture) you can see the arc $s=A\theta$ and you also can see the when $\lim_{\theta\to0}H= A,O= s$, so: $$\lim_{\theta\to0}\sin(\theta)=\frac{O}{H}= \frac{O}{A}=\frac{s}{A}=\frac{A\theta}{A}=\theta\\\therefore\lim_{\theta\to0}\sin(\theta)=\theta\implies\lim_{\theta\to0}\frac{\sin(\theta)}{\theta}=1$$

way 2:

the second way is using what Taylor told us:$$\sin(\theta)=\theta -{\frac {\theta ^{3}}{3!}}+{\frac {\theta ^{5}}{5!}}-{\frac {\theta ^{7}}{7!}}+\cdots$$ now $0<n<1\implies n^m>n^{m-1}$ where $m\in\Bbb N$

that fact shows as that as $\lim_{\theta\to0}\theta^m$ getting even smaller, $\theta$ is positive so $\theta^m$ is also positive, so we can conclude that as $\lim_{\theta\to0}$ $\theta^m$ approaches $0$ exponentially faster so$$\lim_{\theta\to0}\sin(\theta)=\theta -\underbrace{{\frac {\theta ^{3}}{3!}}+{\frac {\theta ^{5}}{5!}}-{\frac {\theta ^{7}}{7!}}+\cdots}_{\text{exponentially approaching 0}}= \theta\\\therefore\lim_{\theta\to0}\sin(\theta)=\theta\implies\lim_{\theta\to0}\frac{\sin(\theta)}{\theta}=1$$

ℋolo
  • 10,006
  • 1
    The limits are exact and should be $=$, not $\approx$. Also rather than saying $\lim_{\theta\to0}\sin(\theta)=\lim_{\theta\to0}\theta$, it would be better to just divide the Taylor series by $\theta$, and find $$\lim_{\theta\to0}\frac{\sin(\theta)}{\theta}=\lim_{\theta\to0}\left(1-\frac{\theta^2}{3!}+\frac{\theta^4}{5!}-...\right)=1$$ directly like this, with all but the first term being equal to $0$. – John Doe Sep 28 '17 at 19:48
  • @JohnDoe i changed the $\approx$ to $=$ but i think that understanding that powers have power also in small numbers and not only large numbers is very important, especially if you are not used to work with those, so i prefer how i showed it – ℋolo Sep 28 '17 at 20:04
  • Yep, that is true. Even so, I will leave my comment there to show an alternative way of computing the same limit. – John Doe Sep 28 '17 at 20:06
  • What you wrote does not appear to make sense mathematically. Thus $\lim_{\theta\to 0}\sin\theta =0$ and not equal to $\theta$. Anyway the first proof via geometrical arguments is not at all rigorous and therefore incorrect. It is precisely these type of hand waving arguments which must be avoided if one wants to have a serious understanding of calculus. Your second proof also suffers from similar flaws. – Paramanand Singh Sep 29 '17 at 02:51