Fundumental Trig Limit

[ChaoticLlama]

A question was posed by my calculus teacher.

We had proven that when using radian measure, the fundumental trig limit of

\(\displaystyle {\lim }\limits_{x \to 0} \frac{{\sin (x)}}{x} = 1\)

However, the question he gave us was why when using degree measure the limit changes to

\(\displaystyle {\lim }\limits_{x \to 0} \frac{{\sin (x)}}{x} \approx 0.01745329\)

Any explanations why?

I was thinking that it has something to do with the tangent approximation of sin(x) = x for x near 0. But that only further supports the original proven limit, not a reason for the other limit to differ.

 

[galactus]

Because \(\displaystyle \frac{\pi}{180}=.01745329252\)

It's that radian and degree thing.

 

[soroban]

Hello, ChaoticLlama!

Galactus nailed it . . . That is <u>exactly</u> the reason!

We had proven that when using radian measure, the fundumental trig limit of

\(\displaystyle {\lim }\limits_{z\to 0} \frac{{\sin (z)}}{z}\,=\,1\)

However, the question he gave us was why when using degree measure the limit changes to

\(\displaystyle {\lim }\limits_{x \to 0} \frac{{\sin (x)}}{x}\,\approx\,0.01745329\)

Any explanations why?

Note that in the original theorem, the three \(\displaystyle z\)'s must be identical.

If \(\displaystyle x\) is in degrees, we can convert it to radians: \(\displaystyle \,\frac{\pi x}{180}\)

So we have: \(\displaystyle \L\;\frac{\sin\left(\frac{\pi x}{180}\right)}{x}\)
Multiply top and bottom by \(\displaystyle \frac{\pi}{180}:\;\;\L\;\;\;\frac{\frac{\pi}{180}\cdot\sin\left(\frac{\pi x}{180}\right)}{\frac{\pi x}{180}\)

Note that, if \(\displaystyle x\to0\), then \(\displaystyle \frac{\pi x}{180} \to 0\)

Let \(\displaystyle z\,=\,\frac{\pi x}{180}\)

Then we have: \(\displaystyle \L\;\lim_{z\to0}\left[\frac{\pi}{180}\cdot\frac{\sin(z)}{z}\right]\;=\;\frac{\pi}{180}\cdot\lim_{z\to0}\left[\frac{\sin(z)}{z}\right]\;=\;\frac{\pi}{180}\cdot1\;=\;\frac{\pi}{180}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This type of acrobatics comes up in simpler problems . . .

Example: \(\displaystyle \L\:\lim_{x\to0}\frac{\sin(3x)}{x}\)

Right now, the "three \(\displaystyle z\)'s" are not the same.

Multiply top and bottom by 3:\(\displaystyle \:3\cdot\frac{\sin(3x)}{3x}\)

Then note that, if \(\displaystyle x\to0\), then \(\displaystyle 3x\to0\)

So we have: \(\displaystyle \L\:\lim_{3x\to0}\left[3\cdot\frac{\sin(3x)}{3x}\right] \;= \;3\cdot\left[\lim_{3x\to0}\frac{\sin(3x)}{3x}\right]\)

Now the \(\displaystyle z\)'s are the same; we can apply the theorem:

\(\displaystyle \L\;\;\;3\cdot\left[\lim_{z\to0}\frac{\sin(z)}{z}\right]\;=\;3\cdot1\;=\;3\)

 

[ChaoticLlama]

thanks for your help

 

[galactus]

Thanks Soroban for the enthusiastic agreement. That's reassuring.

Anyway, ChaoticLama, sometime try doing a simple trig integral using degrees and see what a 'wacky' answer you get.

The thing is, radians are an absolute. There are 2\(\displaystyle {\pi}\) radians in a circle. But the fact there are 360 degrees in a circle is how we have chosen to break it up(from the Babylonians, I believe). Conceivably, there could be any number of degrees in a circle. During WWII, they used gradients(which you still see on some calculators); there were 400 degrees in a circle because they thought GI's could understand better.