Is it wrong? (A theorem of limits)

[Adit]

In attachments, there is a pic of a page, I think there area of sector MOA should be 2π^2/x. Please tell me how this makes sense?

 

[pat1enc3_17]

I would solve this with L’Hospital's rule

\begin{align}
\frac{f(x)}{g(x)}:= \frac{sin(x)}{x}\qquad \text{ since}\quad g'(x)\neq 0,
\end{align}
it follows with L’Hospital's rule
\begin{align}
\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{f'(x)}{g'(x)} =\lim_{x\rightarrow 0} \frac{cos(x)}{1}=1,
\end{align}

 

[gopher_p]

View attachment 71628 In attachments, there is a pic of a page, I think there area of sector MOA should be 2π^2/x. Please tell me how this makes sense?

The area of a circular sector of radius ##r## spanned by an angle ##\theta## is ##\frac{1}{2}r^2\theta##. Alternatively (and used by this text), the area is ##\frac{1}{2}rL##, where ##L=r\theta## is the length of the arc of the circle of radius ##r## spanned by the angle ##\theta##.

I would solve this with L’Hospital's rule

\begin{align}
\frac{f(x)}{g(x)}:= \frac{sin(x)}{x}\qquad \text{ since}\quad g'(x)\neq 0,
\end{align}
it follows with L’Hospital's rule
\begin{align}
\lim_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 0}\frac{f'(x)}{g'(x)} =\lim_{x\rightarrow 0} \frac{cos(x)}{1}=1,
\end{align}

One usually uses the fact that ##\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}=1## to establish ##\frac{d}{dx}\sin x=\cos x##. Using ##\frac{d}{dx}\sin x=\cos x## in the process of using l'Hopital's rule to show that ##\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}=1## is a circular argument.

 

[Erland]

For a full circle, the centre angle is 2π radians (one full lap), and the area is πR². For a sector of a circle, it is clear that the area A is proportional to the the centre angle θ.
Since θ=2π corresponds to A=πR², it follows that in general: θ/2π = A/πR², which gives A=θR²/2. In particular: if R=1, A=θ/2.
Also, by the definition of a radian, if x is the length of the circle sector, θ=x (if R=1).

Using L'Hôpital's rule to calculate the limit ##\lim_{x\to 0}\frac{\sin x}x## is no good idea, because it would probably be circular: this limit is in most cases used in the calculation of the derivative of ##\sin x##.

 

[HallsofIvy]

One usually uses the fact that ##\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}=1## to establish ##\frac{d}{dx}\sin x=\cos x##. Using ##\frac{d}{dx}\sin x=\cos x## in the process of using l'Hopital's rule to show that ##\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}=1## is a circular argument.

Usually, but not necessarily. There are other ways of defining "sine" and "cosine" that do not require a geometric way of getting the derivative of sin(x) and cos(x).

For example, we can define sin(x) to be the power series
[tex]\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}[/tex]
and define cos(x) to be the power series
[tex]\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}[/tex]

Those series have infinite radius of convergent so converge uniformly on any finite interval and, in particular are differentiable "term by term" at any x. Differentiating the series for sin(x) term by term gives
[tex]\sum_{n= 0}^\infty\frac{(2n+1)x^{2n}}{(2n+1)!}= \sum_{n= 0}^\infty\frac{x^{2n}}{(2n)!}= cos(x)[/tex].


Another way to define sine and cosine are as solutions to an 'initial value problem':

Define cos(x) to be the function, y(x), satisfying the differential equation y''= -y with the initial conditions y(0)= 1, y'(0)= 0,

Define sin(x) to be the function, y(x), satisfying the differential equation y''= -y with the initial conditions y(0)= 0, y'(0)= 1.

By the fundamental "existence and uniqueness" theorem for differential equations, there exist unique functions satisfying those. In particular, if y is the derivative of sine, if y(x)= (sin(x))' then y'(x)= (sin(x))''= -(sin(x)) so that y''(x)= -(sin(x))'= -y. That is, y(x)= (sin(x))' satisfies the same differential equation. Further, y(0)= (sin(x))' at 0 which is 1. y'(0)= -(sin(0))= 0. That is, the derivative of sin(x) is cos(x).

 

[gopher_p]

Usually, but not necessarily. There are other ways of defining "sine" and "cosine" that do not require a geometric way of getting the derivative of sin(x) and cos(x).

For example, we can define sin(x) to be the power series
[tex]\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}[/tex]
and define cos(x) to be the power series
[tex]\sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}[/tex]

Those series have infinite radius of convergent so converge uniformly on any finite interval and, in particular are differentiable "term by term" at any x. Differentiating the series for sin(x) term by term gives
[tex]\sum_{n= 0}^\infty\frac{(2n+1)x^{2n}}{(2n+1)!}= \sum_{n= 0}^\infty\frac{x^{2n}}{(2n)!}= cos(x)[/tex].


Another way to define sine and cosine are as solutions to an 'initial value problem':

Define cos(x) to be the function, y(x), satisfying the differential equation y''= -y with the initial conditions y(0)= 1, y'(0)= 0,

Define sin(x) to be the function, y(x), satisfying the differential equation y''= -y with the initial conditions y(0)= 0, y'(0)= 1.

By the fundamental "existence and uniqueness" theorem for differential equations, there exist unique functions satisfying those. In particular, if y is the derivative of sine, if y(x)= (sin(x))' then y'(x)= (sin(x))''= -(sin(x)) so that y''(x)= -(sin(x))'= -y. That is, y(x)= (sin(x))' satisfies the same differential equation. Further, y(0)= (sin(x))' at 0 which is 1. y'(0)= -(sin(0))= 0. That is, the derivative of sin(x) is cos(x).

If we use the power series definition, then $$\frac{\sin x}{x}=\sum_{n=0}^\infty \frac{x^{2n}}{(2n+1)!}$$ for all ##x\neq 0##, and the limit can be solved by direct substitution.

If we use the DE definition, then, for ##y(x)=\sin x## we get $$\lim_{x\rightarrow 0}\frac{\sin x}{x}=\lim_{x\rightarrow 0}\frac{y(x)-y(0)}{x-0}=y'(0)=1$$ for free using the definition of the derivative.

So the use of l'Hopital's Rule to solve the limits given either of those definitions, while not circular, is a bit ham-handed. Though I guess I'll need to rephrase my comment ...

It's bad pedagogy to advise students to use l'Hopital's rule to evaluate ##\lim\limits_{x\rightarrow 0}\frac{\sin x}{x}##. It's especially bad when you have been reminded that that vast majority of students asking about that limit are (a) learning about that limit so that they can turn around and use it to prove things about the derivatives of trig functions and (b) likely weeks away from even hearing about l'Hopital's rule.