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-   -   I can't show why x->0 of sin(x)/x is 1!!! (http://mymathforum.com/calculus/346708-i-cant-show-why-x-0-sin-x-x-1-a.html)

 babaliaris July 5th, 2019 02:23 PM

I can't show why x->0 of sin(x)/x is 1!!!

I started using $\displaystyle -|x| \leq \sin x \leq |x|$ then $\displaystyle -\frac{|x|}{x} \leq \frac{\sin x}{x} \leq \frac{|x|}{x}$

But there is not a single value L for $\displaystyle \lim_{x \to 0} \frac{|x|}{x}=L$ .

Well I have a theory but I'm not sure if it stands.
$\displaystyle -\lim_{x \to 0^-}\frac{|x|}{x} = -(-1) = 1$

$\displaystyle \lim_{x \to 0^+}\frac{|x|}{x} = 1$

So $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}=1$

But can I actually do that? I took different kind of limits in both cases.
This seems like I'm breaking the inequality since I'm taking different limits at once! It feels like I'm doing something like this: $\displaystyle x + 5 = 5 <=> (x + 5) \cdot 10 = 5 \cdot 20$ I multiplied the two sides of the equation with not the same number.

 romsek July 5th, 2019 02:46 PM

https://www.math.tamu.edu/~mpilant/m.../solutions.pdf

 Greens July 5th, 2019 06:07 PM

The Calculus way to go about this would be L'hopitals rule:

If $\lim_{x \to a} \frac{f(x)}{g(x)}$ is indeterminate ($\frac{0}{0}$, $\frac{\infty}{\infty}$), then

$\displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}$

So, since $\lim_{x \to 0} \frac{\sin{x}}{x} = \frac{0}{0}$

$\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x} = \lim_{x \to 0}\frac{\cos{x}}{1} = \cos{0}=1$

But the method Romsek sent is a much more concrete way to go about it.

 skipjack July 5th, 2019 10:03 PM

Quote:
 Originally Posted by Greens (Post 611354) The Calculus way to go about this . . .
You assume that the derivative of sin(x) is cos(x), but the usual proof of that assumes knowledge of the limit you're trying to find.

 Greens July 5th, 2019 10:37 PM

That's absolutely true, which is why I mentioned the other method is much more concrete.

But generally, if I were tossed this problem without having to build from the ground up I would use L'hôpital.

 babaliaris July 6th, 2019 12:24 AM

Quote:
 Originally Posted by Greens (Post 611354) The Calculus way to go about this would be L'hopitals rule: If $\lim_{x \to a} \frac{f(x)}{g(x)}$ is indeterminate ($\frac{0}{0}$, $\frac{\infty}{\infty}$), then $\displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}$ So, since $\lim_{x \to 0} \frac{\sin{x}}{x} = \frac{0}{0}$ $\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x} = \lim_{x \to 0}\frac{\cos{x}}{1} = \cos{0}=1$ But the method Romsek sent is a much more concrete way to go about it.
I tried to understand Romsek's answer but I could not understand the math. Since I'm re-studying calculus, I perfectly understand your answer even currently in the book I haven't even reach the derivatives chapter (I'm still on limits and now entering continuity).

Also my original post of taking the left handed side and the right handed side limit is wrong write?
−limx→0−|x|x=−(−1)=1

limx→0+|x|x=1

So limx→0sinxx=1

 Greens July 6th, 2019 06:34 AM

When you took $-\lim_{x \to 0^{-}} \frac{|x|}{x}$ , the added negative sign breaks it since the fraction is already negative on the left. This creates the squeeze

$\displaystyle \frac{|x|}{x} \leq \frac{\sin{x}}{x} \leq \frac{|x|}{x}$

Which would imply $\frac{|x|}{x} = \sin{x}$ , which isn't always true.

 SDK July 6th, 2019 01:53 PM

So much is wrong in this post which isn't really worth trying to correct. Two things are bad enough they should be pointed out.

1. You absolutely can NOT use L'Hospital's rule to "show" that $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$. At best this is circular logic, but I would say it's even worse than that. It is a complete misunderstanding of what L'Hospital's rule means.

2. Romsek's link is technically accurate, but I don't think it's helpful. It begins with 2 inequalities which "can be shown". But given these 2 inequalities it's completely trivial to compute this limit. The important part (which also has all of the insight here) is showing those two inequalities. This is typically done by a geometric argument which is difficult to carry out without drawing the right picture.

 romsek July 6th, 2019 02:09 PM

Quote:
 Originally Posted by SDK (Post 611375) So much is wrong in this post which isn't really worth trying to correct. Two things are bad enough they should be pointed out.
thank you so much for stopping by to criticize what's been posted.

I don't suppose it would be too much to ask that you post what you consider an adequate solution to the problem being asked would it?

 babaliaris July 8th, 2019 05:29 AM

Quote:
 Originally Posted by SDK (Post 611375) So much is wrong in this post which isn't really worth trying to correct. Two things are bad enough they should be pointed out. 1. You absolutely can NOT use L'Hospital's rule to "show" that $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$. At best this is circular logic, but I would say it's even worse than that. It is a complete misunderstanding of what L'Hospital's rule means. 2. Romsek's link is technically accurate, but I don't think it's helpful. It begins with 2 inequalities which "can be shown". But given these 2 inequalities it's completely trivial to compute this limit. The important part (which also has all of the insight here) is showing those two inequalities. This is typically done by a geometric argument which is difficult to carry out without drawing the right picture.
Can someone explain why I can't use L'Hospital's rule here? Are you sure about it? From what I remember from high school, if you have a limit of the form $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$ you can use L'Hospital's rule.

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