$\begingroup$ You can't calculate exact value of sin(x)/x for x=$0$. When you say x tends to $0$, you're already taking an approximation.So, we have to calculate the limit here.Taylor series gives very accurate approximation of sin(x), so it can be used to calculate limit. $\endgroup$

$\begingroup$ The purposes of the question you linked to is to precisely answer why "We know that for small values of x that sinx is approximately equal to x"? $\endgroup$ – user17762 Commented Feb 24, 2013 at 0:23

How can one prove the statement $$\lim_{x\to 0}\frac{\sin x}x=1$$ without using the Taylor series of $\sin$, $\cos$ and $\tan$?

$\begingroup$ Are you sure lim[(sinx)/x] = 0 when x aproaches infinity ? I mean it's obvious it's 0 because you divide a number between -1 and +1 with something that approaches infinity, but we now study limits and we were not told that. $\endgroup$

2018年1月19日 · I knew that if I show that each limit was 1, then the entire limit was 1. I decided to start with the left-hand limit. For x<0, 1/x <= sin(x)/x <= -1/x. However, since the limit as x approaches 0 from the left of 1/x = -oo and the limit as x approaches 0 from the left of -1/x is oo, the squeeze theorem really can't be applied.

2017年1月29日 · I have heard people say that you can't (or shouldn't) use the L'Hopital's rule to calculate $\lim\limits_{x\to 0}\frac{\sin x}x=\lim\limits_{x\to 0}\cos x=1$, because the result $\frac d{dx}\sin x=\cos x$ is derived by using that limit.

2015年12月27日 · I'm trying to find a proof that: $$ \lim_{x\to0^-}\sin(x)=0$$ I'd like to be able to do the proof without reference to advanced theorems (mean value theorem, series, etc). I have a geometric approach for finding the limit from the right, but I need similar help when approaching zero from the left.

$\begingroup$ It seems to me that there is a big problem with using the Taylor series. Notice that $$\frac{d}{dx} \sin x := \lim_{h \to 0} \frac{\sin(x+h)-\sin x}{h} \equiv \lim_{h \to 0} \left[ \left(\frac{\cos h -1}{h}\right) \sin x+ \left(\frac{\sin h}{h}\right) \cos x \right].$$ By using the Taylor series, you are using the fact that the derivative of $\sin x$ is $\cos x$, and so are ...

2019年9月13日 · Lim to 0, sinx/x theory. Ask Question Asked 5 years, 3 months ago. Modified 5 years, 3 months ago. Viewed ...

2017年5月23日 · Can you please help me solve: $$\lim_{x \rightarrow 0} \frac{1- \cos x}{x \sin x}$$ Every time I try to ...