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Is it really circular to prove the sinx/x limit with L'Hopital? 
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#1
Oct2812, 11:41 AM

P: 427

Hello,
This limit [tex]\lim_{x \to 0} \frac{ \sin x}{x}[/tex] is often cited as being an example where L'Hopital's rule cannot be used, since to use it you'd need to differentiate sine; but the derivative of sine, using the limit definition of a derivative, requires that you use the sinx/x limit (and the 1  cosx / x limit) as part of the proof. So, we'd end up with a circular proof here, and thus we'd have to use the squeeze theorem as the alternative. But what struck me is if that is really the only proof that cosine is the derivative of sine? Aren't there others? Apologies if this topic has been done to death. 


#2
Oct2812, 12:28 PM

Mentor
P: 18,038

Everything depends on your definition of sine and cosine. If you define it as an integral or as a series, then you don't need to know the sin(x)/x limit.



#3
Oct2812, 04:55 PM

HW Helper
P: 2,263

First L'Hopital would not be needed as it is stronger than needed. That limit is sin'(0). As micromass said just define sine so that you can find sin'(0) and there is no problem. I just include sin'(0)=1 in the definition, problem solved. Any definition of sine will lead us to the fact that sin'(0) needs to exist and not be zero. We prefer sin'(0)=1 for simplicity though we often hide this condition as
sin(x)~x (x small) sin'(x)=cos(x) 4Arcsin(sqrt(2)/2)=pi but all these are equivalent to sin'(0)=1. 


#4
Dec812, 02:53 AM

P: 47

Is it really circular to prove the sinx/x limit with L'Hopital?
I don't know why high school students keep trying to use L'Hopital, these kinds of limits are really simple, my tutor taught me a method and it works every time (at least at high school level).
For example; [tex]\lim_{x\rightarrow 0}\frac{sin3x}{5x}=\lim_{x\rightarrow 0}\frac{sin3x}{3x}\times \frac{3}{5}=\frac{3}{5}[/tex] or something harder like; [tex]\lim_{x\rightarrow 0}\frac{sinx}{x(1+cosx)}=\lim_{x\rightarrow 0}\frac{2sin\frac{x}{2}cos\frac{x}{2}}{2xcos^{2}(x/2)}=\lim_{x\rightarrow 0}\frac{tan\frac{x}{2}}{\frac{x}{2}}\times \frac{1}{2}=\frac{1}{2}[/tex] 


#5
Dec812, 09:34 AM

Sci Advisor
P: 1,716

I would try a geometric proof of this limit if you want to do it from scratch. 


#6
Dec812, 09:53 AM

Sci Advisor
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PF Gold
P: 12,016

You meet the limit of sin(x)/x very early on in your studies.
It is perfectly possible to define sin(x) in such a way that application of L'Hopital won't become circular, but usually it will be, since, for example, regarding sin(x) as defined as the solution of an eigenvalue problem won't be the definition an early calculus student is equipped with. Neither will an infinite series definition be the usual case.  Thus, a GEOMETRIC derivation of the limit will typically be the one noncircular proof left to the fledgling student.  The simplest of those, in my view, is by a squeeze theorem application based on area comparisons on the unit circle, utilizing the radian measurement of angles. An inscribed rightangled triangle has sides cos(x), sin(x), and 1, giving area cos(x)sin(x)/2. The circular sector has area x/2 And the rightangled triangle just circumscribing that has sides 1, tan(x), 1/sec(x), with area tan(x)/2.  the limit of sin(x)/x follows easily from this, remembering that cos(x) tends to 1 as x goes to 0. 


#7
Dec812, 09:47 PM

HW Helper
P: 2,263

Some of the less careful geometric proofs are circular. The key point is that
lim sin(x)/x should exist and not be zero and the scaling where it is one is chosen for convenience. 


#8
Dec912, 12:11 AM

P: 783

BiP 


#9
Dec912, 10:56 AM

Sci Advisor
P: 1,716

One could also start with the power series definition of the sine and differentiate it term by term to get the cosine. This avoids using the Newton quotient sin(x)/x or also use various other definition of sine such as (e[itex]^{ix}[/itex]  e[itex]^{ix}[/itex])/2i
I never tried this but maybe you can differentiate the continued fraction definition of the sine. 


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