4
$\begingroup$

The full proof can be found here. Basically, we compare the three areas that depend on $x$ in the circle of radius $1$ shown below.

enter image description here

Regardless of the value of $x$, we should have

$$\text{area of sector OAC} < \text{area of triangle OAP} < \text{area of sector OBP}$$ $$\frac{1}{2}x(\cos{x})^2<\frac{1}{2}(\cos{x})(\sin{x})<\frac{x}{2}$$

I have two questions about it:

1) Shouldn't there be a $\le$ instead of the $<$ sign in the inequality above, since the areas of sector $OAC$ and the area of triangle $OAP$ both become zero when $x=\frac{\pi}{2}$?

2) If the value of $x$ is such that we end up in the fourth quadrant, the value of $\sin{x}$ becomes negative and the inequality no longer holds (since $\frac{1}{2}x(\cos{x})^2>0$ and $\frac{1}{2}(\cos{x})(\sin{x})<0$). How can we go around this?

Thanks in advance.

$\endgroup$
3
$\begingroup$

1) When we are talking about $x\to0$, we are looking at those $x$ close to $0$, but not equal to $0$. The value of the function at $x=0$ is just irrelevent. So $<$ is correct.

2) For $\frac{\pi}{2}>x>0$, we have

$$\frac{1}{2}x(\cos{x})^2<\frac{1}{2}(\cos{x})(\sin{x})<\frac{x}{2}$$

This implies that

$$\cos x<\frac{\sin x}{x}<\frac{1}{\cos x}$$

For $\frac{-\pi}{2}<x<0$, let $y=-x$. Then $\frac{\pi}{2}>y>0$ and hence

$$\cos y<\frac{\sin y}{y}<\frac{1}{\cos y}$$

Note that $\sin x=\sin(-y)=-\sin y$ and $\cos x=\cos(-y)=\cos y$.

So we still have

$$\cos x<\frac{\sin x}{x}<\frac{1}{\cos x}$$

$\endgroup$
3
$\begingroup$
  1. Yes, if you allow degenerate figures, like triangles with no area, then you have a weak inequality instead of strict inequality. In fact the usual version of the squeeze theorem requires this to be weak inequality. Though since that limit point is outside the domain of this strict inequality, it's ok that your text uses strict.

  2. at $x=\pi/2$, the formula no longer holds, as the triangle's area is zero, but the circular sectors are not. Just look at your figure.

$\endgroup$
  • $\begingroup$ Why do we need weak inequality for Squeeze theorem? It works with strong inequality also. $\endgroup$ – Paramanand Singh Jan 13 '18 at 6:57
  • $\begingroup$ @ParamanandSingh If the limit point is an element of the domain, then strict inequality is incompatible with the consequent of the theorem. However in the case that the limit point is not contained in the domain, I think you're right, then strict inequality is fine. The statements of the theorem I have seen presentations of always include weak inequalities, but understanding that distinction would give you a stronger theorem. $\endgroup$ – ziggurism Jan 13 '18 at 14:39
  • $\begingroup$ Agree with whatever you said! +1 $\endgroup$ – Paramanand Singh Jan 13 '18 at 14:42
2
$\begingroup$

For question 1)

Note that the limit is taken at $x=0$. The equality occurs at $x= \frac {\pi}{2}$ so removing the equality sign really does not matter.

For question 2)

Upon dividing both sides of $$\frac{1}{2}x(\cos{x})^2<\frac{1}{2}(\cos{x})(\sin{x})<\frac{x}{2}$$by a negative $x$ results in $$\frac{1}{2}(\cos{x})^2>\frac{1}{2}(\cos{x})(\frac {sin{x}}{x})>\frac{1}{2}$$ Notice that if $x$ approaches $0$ we still get the desired result of $$\lim_{x\to 0} \frac {\sin x}{x}=1$$.

$\endgroup$
1
$\begingroup$

@sirous, You can't use L'Hopital rule to evaluate this limit. It results in Circular Reasoning because how did u know that (sinx)' = cosx? It requires a knowledge of the desired limit to evaluate. Hence using L'Hopital Rule results in a mathematical fallacy. The best answers have been stated above.

$\endgroup$
  • $\begingroup$ L'H is valid because $\lim_{x \rightarrow 0}sin(x) = 0$ and $\lim_{x \rightarrow 0} x = 0$. It's a $\frac{0}{0}$ indeterminant form which is sufficient for L'H. $\endgroup$ – Quelklef Jan 13 '18 at 6:58
  • $\begingroup$ I wish. It does satisfy the conditions of L'H, I have no problem with that. The thing is; how did u know how (sinx)' = cosx? Can you prove this without using the desired limit to evaluate? $\endgroup$ – user517110 Jan 13 '18 at 7:02
  • $\begingroup$ Prove $\lim_{x \rightarrow 0}{\frac{\sin x}{x}}$ another way, use that to prove $\frac{d}{dx}\sin x$, use that to prove $\lim_{x \rightarrow 0}{\frac{\sin x}{x}}$ this way. Since the discussion is about the validity of this proof specifically, rather than its conclusion, this will work. $\endgroup$ – Quelklef Jan 13 '18 at 17:20
  • $\begingroup$ In case you don't like that alternative proof. $\endgroup$ – Quelklef Jan 13 '18 at 17:25
1
$\begingroup$

Your proof should first explain that the function $(\sin x) /x$ is even and hence it is sufficient to consider the limit as $x\to 0^{+}$ and thus we can take $x>0$. Moreover the limit is dependent on behavior of function in arbitrarily small neighborhood of $0$ and thus one can take the interval $(0,\pi/2)$ as the range of values of $x$ to deal with this limit evaluation.

$\endgroup$
1
$\begingroup$

$$\lim_{x\to0}\frac{\sin(x)}{x}=\lim_{x\to0}\left(\frac {(\sin(x))'}{x'}\right)=\left(\frac{\cos(x)}{1}\right)\large|_{x=0}=1$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.