What algebraic manipulation did they do to get this answer?

[warwick]

http://calcchat.tdlc.com/free_solutions/main.html

Textbook: Calculus, 8th Edition
Chapter: 1
Section: R
Exercise: 23

I see how they initially expanded everything, but where did the -1 come from in the (cos delta(x) - 1) and what happened to the (-1/2)?

 

[galactus]

Perhaps you can post the problem. I went to the site and it takes too long for it to load on my old, dinosaur dial up.

 

[warwick]

Perhaps you can post the problem. I went to the site and it takes too long for it to load on my old, dinosaur dial up.

 

[galactus]

Notice, this is the definition of a derivative.

You can use L'Hopital's rule. Since this is an indeterminate form, 0/0

I will use h instead of delta(x)

\(\displaystyle \L\\\lim_{h\to\0}\frac{sin(\frac{\pi}{6}+h)-\frac{1}{2}}{h}\)

Differentiate top and bottom:

\(\displaystyle \L\\\lim_{h\to\0}cos(\frac{\pi}{6}+h)\)

Of course, now you can see the limit is \(\displaystyle cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\)

What we did was find the derivative of sin(x), then evaluated at Pi/6.

Is this what they were doing that you wondered where the cosine came from?.

 

[warwick]

Notice, this is the definition of a derivative.

You can use L'Hopital's rule. Since this is an indeterminate form, 0/0

I will use h instead of delta(x)

\(\displaystyle \L\\\lim_{h\to\0}\frac{sin(\frac{\pi}{6}+h)-\frac{1}{2}}{h}\)

Differentiate top and bottom:

\(\displaystyle \L\\\lim_{h\to\0}cos(\frac{\pi}{6}+h)\)

Of course, now you can see the limit is \(\displaystyle cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\)

What we did was find the derivative of sin(x), then evaluated at Pi/6.

Is this what they were doing that you wondered where the cosine came from?.

Well, I started reviewing my Cal I chapters since I took it last fall, so I was trying not to apply any calculus techniques to determine the limit. I can post a picture of the way the book worked it. Give me a minute.

 

[warwick]

Notice, this is the definition of a derivative.

You can use L'Hopital's rule. Since this is an indeterminate form, 0/0

I will use h instead of delta(x)

\(\displaystyle \L\\\lim_{h\to\0}\frac{sin(\frac{\pi}{6}+h)-\frac{1}{2}}{h}\)

Differentiate top and bottom:

\(\displaystyle \L\\\lim_{h\to\0}cos(\frac{\pi}{6}+h)\)

Of course, now you can see the limit is \(\displaystyle cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\)

What we did was find the derivative of sin(x), then evaluated at Pi/6.

Is this what they were doing that you wondered where the cosine came from?.

Well, I started reviewing my Cal I chapters since I took it last fall, so I was trying not to apply any calculus techniques to determine the limit. I can post a picture of the way the book worked it. Give me a minute.

 

[galactus]

Yes, I was just coming back to say another way to do this is by using the addiiton formula for \(\displaystyle sin(\frac{\pi}{6}+h)\)

That's what they've done here.

The addition formula you may remember. It's

\(\displaystyle \L\\sin(u+v)=sin(u)cos(v)+cos(u)sin(v)\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{sin(\frac{\pi}{6})cos(h)+cos(\frac{\pi}{6})sin(h)-\frac{1}{2}}{h}\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{\frac{1}{2}cos(h)+\frac{\sqrt{3}}{2}sin(h)-\frac{1}{2}}{h}\)

Now, factor out 1/2:

\(\displaystyle \L\\\lim_{h\to\0}\frac{\frac{1}{2}(cos(h)-1)+\frac{\sqrt{3}}{2}sin(h)}{h}\)

\(\displaystyle \L\\\frac{1}{2}\underbrace{\lim_{h\to\0}\frac{cos(h)-1}{h}}_{\text{limit=0}}+\frac{\sqrt{3}}{2}\underbrace{\lim_{h\to\0}\frac{sin(h)}{h}}_{\text{limit=1}}\)

Therefore, the limit is \(\displaystyle \L\\\frac{\sqrt{3}}{2}\)

See how that works?.

I should've posted it that way. I always like to do limits without L'Hopital if possible. It was less typing.

 

[warwick]

Yes, I was just coming back to say another way to do this is by using the addiiton formula for \(\displaystyle sin(\frac{\pi}{6}+h)\)

That's what they've done here.

The addition formula you may remember. It's

\(\displaystyle \L\\sin(u+v)=sin(u)cos(v)+cos(u)sin(v)\)

I should've posted it that way. I always like to do limits without L'Hopital if possible.

I guess I'm missing some the implied algebraic manipulation in their steps. Refer back to my original question, please.

 

[soroban]

Re: What algebraic manipulation did they do to get this answ

Hello, warwick!

You are expected to know these two theorems:

. . \(\displaystyle \lim_{\theta\to 0}\frac{1\,-\,\cos\theta}{\theta} \:=\:0\;\;\;\;\;\lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1\)

\(\displaystyle \L\lim_{\Delta x\to0}\frac{\sin(\frac{\pi}{6}\,+\,\Delta x)\,-\,\frac{1}{2}}{\Delta x}\)

Using a compound-angle identity, we have:

. . \(\displaystyle \L\frac{\sin(\frac{\pi}{6})\cos(\Delta x)\,+\,\sin(\Delta x)\cos(\frac{\pi}{6})\,-\,\frac{1}{2}}{\Delta x} \;=\;\frac{\frac{1}{2}\cos(\Delta x)\,+\,\frac{\sqrt{3}}{2}\sin(\Delta x)\,-\,\frac{1}{2}}{\Delta x}\)

. . \(\displaystyle \L=\;\frac{1}{2}\left[\frac{\cos(\Delta x)\,-\,1\,+\,\sqrt{3}\sin(\Delta x)}{\Delta x}\right] \;=\;\frac{1}{2}\left[\sqrt{3}\cdot\frac{\sin(\Delta x)}{\Delta x}\,-\,\frac{1\,-\,\cos(\Delta x)}{\Delta x}\right]\)


Then we have: \(\displaystyle \L\:\lim_{\Delta x\to0}\left[\frac{1}{2}\left(\sqrt{3}\cdot\frac{\sin(\Delta x)}{\Delta x}\,-\,\frac{1\,-\,\cos(\Delta x)}{\Delta x}\right)\right]\)

. . \(\displaystyle \L=\;\frac{1}{2}\left[\sqrt{3}\cdot\lim_{\Delta x\to0}\left(\frac{\sin(\Delta x)}{\Delta x}\right) \:+\:\lim_{\Delta x\to0}\left(\frac{1\,-\,\cos(\Delta x)}{\Delta x}\right)\right]\)

. . \(\displaystyle \L=\;\frac{1}{2}\left[\sqrt{3}\cdot1\,-\,0\right] \;=\;\fbox{\frac{\sqrt{3}}{2}}\)

 

[tkhunny]

Come on, you must show a little fortitude.

Break up that giant fraction into three pieces. Like this:

(A + B + C)/D = (A/D)+(B/D)+(C/D)

Rearrange a few things and you will find it. The 1/2 didn't go anywhere. Track it down.

 

[warwick]

Ah, sorry, guys. Usually, if I just write everything out I can easily find the solution. Sometimes, though, I try to skip ahead, but the answer typically becomes more obscure that way. I learn best if I see EVERY little detail worked in a solution. Plus, I'm just getting back into the groove of Calculus. Heh. Thanks for all your help.

Oh, by the way, I'm not going to learn L'Hopital until Cal II, this Fall semester. Is that unusual/the norm?

 

[galactus]

Yes, I was just coming back to say another way to do this is by using the addiiton formula for \(\displaystyle sin(\frac{\pi}{6}+h)\)

That's what they've done here.

The addition formula you may remember. It's

\(\displaystyle \L\\sin(u+v)=sin(u)cos(v)+cos(u)sin(v)\)

I should've posted it that way. I always like to do limits without L'Hopital if possible.

I guess I'm missing some the implied algebraic manipulation in their steps. Refer back to my original question, please.

Refer back to my edited post, as well as Soroban's. They're the same thing, only different.

 

[warwick]

Yes, I was just coming back to say another way to do this is by using the addiiton formula for \(\displaystyle sin(\frac{\pi}{6}+h)\)

That's what they've done here.

The addition formula you may remember. It's

\(\displaystyle \L\\sin(u+v)=sin(u)cos(v)+cos(u)sin(v)\)

I should've posted it that way. I always like to do limits without L'Hopital if possible.

I guess I'm missing some the implied algebraic manipulation in their steps. Refer back to my original question, please.

Refer back to my edited post, as well as Soroban's. They're the same thing, only different.

Factoring the 1/2 out threw me off. Like I said, this was my first day getting back into the mathematical "groove." Haha. Cut me some slack!

 

[warwick]

Yes, I was just coming back to say another way to do this is by using the addiiton formula for \(\displaystyle sin(\frac{\pi}{6}+h)\)

That's what they've done here.

The addition formula you may remember. It's

\(\displaystyle \L\\sin(u+v)=sin(u)cos(v)+cos(u)sin(v)\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{sin(\frac{\pi}{6})cos(h)+cos(\frac{\pi}{6})sin(h)-\frac{1}{2}}{h}\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{\frac{1}{2}cos(h)+\frac{\sqrt{3}}{2}sin(h)-\frac{1}{2}}{h}\)

Now, factor out 1/2:

\(\displaystyle \L\\\lim_{h\to\0}\frac{\frac{1}{2}(cos(h)-1)+\frac{\sqrt{3}}{2}sin(h)}{h}\)

\(\displaystyle \L\\\frac{1}{2}\underbrace{\lim_{h\to\0}\frac{cos(h)-1}{h}}_{\text{limit=0}}+\frac{\sqrt{3}}{2}\underbrace{\lim_{h\to\0}\frac{sin(h)}{h}}_{\text{limit=1}}\)

Therefore, the limit is \(\displaystyle \L\\\frac{\sqrt{3}}{2}\)

See how that works?.

I should've posted it that way. I always like to do limits without L'Hopital if possible. It was less typing.

And the limit of ( (cos(h)-1)/h) being 0/0, indeterminate, threw me off as well. I guess that's where L'Hoptial's rule will come into play later for me.

 

[warwick]

Ok. This one?

 

[galactus]

Take a look at even values of x.

The derivative of \(\displaystyle \L\\csc(\frac{{\pi}x}{2})\) is \(\displaystyle \L\\-\frac{\pi}{2}csc(\frac{{\pi}x}{2})cot(\frac{{\pi}x}{2})\)

What happens if you enter, say, x=2

 

[daon]

I vaguely remember that if a function is non-differentiable (at a point), it does not mean it is not continuous (at that point). I think the objective on this is to find when (at what x values) the limit does not exist.

When Sin(\(\displaystyle \pi\)x/2)=0 it forces csc to be undefined. It happens at all even multiples of pi/2. i.e. Its valid for everything but the even integers.

 

[galactus]

Good point, daon. But in this case it is not continuous nor differentiable at x=2n.

 

[warwick]

Remember, I'm still in a pre-derivative chapter - the one right before, actually. So, actually, I want to use "pre-calculus/whatever" to solve this.

 

[galactus]

Then, enter in even values of x. You get an undefined result. Therefore, it's not continuous.

\(\displaystyle \L\\\lim_{x\to\2k}csc(\frac{{\pi}x}{2})=\frac{1}{sin(k{\pi})}\)

Any value of k will result in 1/0. Therefore, discontinuous.


In order for a function to be continuous at a point the following rules must hold:

1. f(c) is defined
2. \(\displaystyle \lim_{x\to\c}f(x)\) exists
3. \(\displaystyle \lim_{x\to\c}f(x)=f(c)\)

Now, if x is odd, that's another matter.

\(\displaystyle \L\\\lim_{x\to\2k+1}csc(\frac{{\pi}x}{2})=\frac{1}{cos(k{\pi})}\)

Any values of k will result in 1 or -1. So, it's continuous.

I hope I explained this well enough.