Understanding the Limits and Derivatives of Sine and Cosine Functions

[Stephanus]

Dear PF Forum,
In previous threads, I have asked about sine and cosine. The answer given by the members/mentors/advisor are very clear. But lengthy. Perhaps these yes/no questions that I can simply remember and not forget it (again).
So here we are
1. if h = 0 then sin(h) = 0
2. if $\lim_{h \to 0}$ then sin(h) ≠ 0
3. If $\lim_{h \to 0}$ then sin(h) has no limit

4. if h = 0 then sin(h)/h is undefined
5. if $\lim_{h \to 0}$ then sin(h)/h = 1
6. If $\lim_{h \to 0}$ then sin(h)/h has limit

7. if h = 0 then Cin(h) = 1
8. If $\lim_{h \to 0}$ then cos(h) ≠ 1
9. If $\lim_{h \to 0}$ then cos(h) has limit
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10. if $\lim_{h \to 0}$ does cos(h) - 1 has limit? If yes then what is the value?
11. if $\lim_{h \to 0}$ does (cos(h) - 1)/h has limit? If yes then what is the value?

And in the derivative of Sin(x)
$\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h} + \cos(x)\frac{\sin(h)}{h}$
$\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h} + \cos(x)$

So if the derivative of Sin(x) is Cos(x) then...
$\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h}$ should be zero. then...
12. $\lim_{h \to 0} \cos(h)-1$ should be zero, then...
13. $\lim_{h \to 0} \cos(h)$ should be 1, then...
8. If $\lim_{h \to 0}$ then cos(h) ≠ 1
It seems number 13 contradicts number 8.
I would be very grateful if someone be sokind to answer me. But before that, could you confirm the yes/no question in number 1 to 9, please.
Thank you very much.

 

[RUber]

Dear PF Forum,
In previous threads, I have asked about sine and cosine. The answer given by the members/mentors/advisor are very clear. But lengthy. Perhaps these yes/no questions that I can simply remember and not forget it (again).
So here we are
1. if h = 0 then sin(h) = 0
2. if $\lim_{h \to 0}$ then sin(h) ≠ 0
3. If $\lim_{h \to 0}$ then sin(h) has no limit

Number 2 and 3 don't make sense.
$\lim_{h \to 0} \sin(h) = 0$ but if h is approaching 0 then sin(h) ≠ 0.

4. if h = 0 then sin(h)/h is undefined
5. if $\lim_{h \to 0}$ then sin(h)/h = 1
6. If $\lim_{h \to 0}$ then sin(h)/h has limit

Similarly, I am not too sure of your notation.
$\lim_{h \to 0}sin(h)/h =1$ is true.
No clue what you mean in number 6. Is this supposed to be a counterpoint to number 3 above?

7. if h = 0 then Cin(h) = 1
8. If $\lim_{h \to 0}$ then cos(h) ≠ 1
9. If $\lim_{h \to 0}$ then cos(h) has limit

For 7, do you mean cos(h)? If so, yes.
Again, for 8 and 9, $\lim_{h \to 0} \cos(h) = 1$ but as h approaches 0, cos(h) approaches but is not equal to 1.
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10. if $\lim_{h \to 0}$ does cos(h) - 1 has limit? If yes then what is the value?
11. if $\lim_{h \to 0}$ does (cos(h) - 1)/h has limit? If yes then what is the value?

$\lim_{h \to 0} (\cos(h) - 1) = \lim_{h \to 0} (\cos(h) )- 1=1 -1$
11 is a good question. Do you know L'Hopital's rule? I would start there.

And in the derivative of Sin(x)
$\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h} + \cos(x)\frac{\sin(h)}{h}$
$\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h} + \cos(x)$

That checks out.

So if the derivative of Sin(x) is Cos(x) then...
$\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h}$ should be zero. then...
12. $\lim_{h \to 0} \cos(h)-1$ should be zero, then...
13. $\lim_{h \to 0} \cos(h)$ should be 1, then...
8. If $\lim_{h \to 0}$ then cos(h) ≠ 1
It seems number 13 contradicts number 8.
I would be very grateful if someone be sokind to answer me. But before that, could you confirm the yes/no question in number 1 to 9, please.
Thank you very much.

Right. number 8 is wrong.

 

[DEvens]

A very long series of "yes-no" questions is not a very efficient way to learn this stuff.

Why don't you try to learn the concepts and the notations? Memorizing a bunch of "if lim h" kinds of things is really going to do nothing but confuse you. Especially when you have them stated quite clumsily.

 

[PeroK]

And, what you are memorising makes no sense. Not least, the derivative of sin is cos, there is no "if" about it. Perhaps more important, your post reveals you have not understood the concept of a limit. You are treating limit h -> 0 as an independent property of h, which is not correct.

This is not the way to do or learn rigorous mathematics, I'm sorry to say.

 

[Drakkith]

Indeed. The best way to learn about limits is probably to grab a calculus textbook and, starting from the front, work your way through the book, answering every problem as you go.

 

[Stephanus]

Concerning number two and three.

2. if $\lim_{h \to 0}$ then sin(h) ≠ 0
3. If $\lim_{h \to 0}$ then sin(h) has no limit

Number 2 and 3 don't make sense.

This is because someone told me that the limit musn't always exist.

Please confirm one more thing.
A. $\lim h{\to 0} \text{ } \frac{6}{h} \rightarrow \text{ infinity}$[..]

No.
The limit doesn't exist. If you don't understand why, read up on left-hand and right-hand limits.

Yes. A limit exists if and only if the left-hand and right-hand limits exist and are equal.

So I am trying to understand this by testing some equations. I choose:
$\lim_{h \to 0}$ then $\sin(h)$ ≠ 0 and
$\lim_{h \to 0}$ then $\sin(h)$ has no limit.
$\lim_{h \to 0}$ then $\cos(h)$ has limit.
Just to make sure if I understand left/right hand correctly.

7. if h = 0 then Cin(h) = 1

For 7, do you mean cos(h)? If so, yes.

What?? That must be the effect of copy-paste.

 

[Stephanus]

$\lim_{h \to 0} \sin(h) = 0$ but if h is approaching 0 then sin(h) ≠ 0.

So what is the different between
A: if $\lim_{h \to 0}$ then $\sin(h) = 0$ and
B: if h = 0 then $\sin(h) = 0$?
Isn't the left/right hand of $\lim_{h \to 0} \sin(h)$ is different, but
The left/right hand of $\lim_{h \to 0} \frac{6h}{h}$ is equal, right.
Or
The left/right hand of $\lim_{h \to 0} \frac{\sin(h)}{h}$ is also equal.
but $\lim_{h \to 0} \sin(h)$ the left/right is different.

 

[Stephanus]

A very long series of "yes-no" questions is not a very efficient way to learn this stuff.

Why don't you try to learn the concepts and the notations? Memorizing a bunch of "if lim h" kinds of things is really going to do nothing but confuse you. Especially when you have them stated quite clumsily.

Thanks DEvens for your respons. It's not the equations that I intend to remember. It's the concept of limit that I try to understand. Here, I choose sine and cosine to study limit.
For example:
If h = 0 then sin(h) = 0
if $\lim_{h \to 0}$ is sin(h) = 0? I think sin(h) ≠ 0. But I need confirmation.

if h = 0 then sin(h)/h is undefined
if $\lim_{h \to 0}$ then sin(h)/h = 1. This what we were taught in high school. Now, I just want to know 'why'

 

[Stephanus]

And, what you are memorising makes no sense. [..] Perhaps more important, your post reveals you have not understood the concept of a limit. You are treating limit h -> 0 as an independent property of h, which is not correct.

This is not the way to do or learn rigorous mathematics, I'm sorry to say.

Yes, today I learned that I don't know the concept of limit. I tought I knew all the way from high school. Thanks for pointing it out to me, so I know what to learn.

So if the derivative of Sin(x) is Cos(x) then...
$\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h}$ should be zero. then...

Not least, the derivative of sin is cos, there is no "if" about it.

Perhaps the word "IF" is the wrong word for mathematic.
I should have written.
Because the derivative of Sin(x) is Cos(x) so the proof is:
$\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h}$ should be zero. then, etc...
But, I'm a computer programmer, I use if ... then ... instinctively.

 

[PeroK]

Yes, today I learned that I don't know the concept of limit. I tought I knew all the way from high school. Thanks for pointing it out to me, so I know what to learn.Perhaps the word "IF" is the wrong word for mathematic.
I should have written.
Because the derivative of Sin(x) is Cos(x) so the proof is:
$\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h}$ should be zero. then, etc...
But, I'm a computer programmer, I use if ... then ... instinctively.

What you have written still makes no mathematical sense. For example, as sin and cos are continuous functions we have:

$\forall x$ $\lim_{h \rightarrow x}sin(h) = sin(x)$ and $\lim_{h \rightarrow x}cos(h) = cos(x)$

There is no issue or mystery about these limits. The more interesting ones are:

$\lim_{h \rightarrow 0} \frac{sin(h)}{h} = 1$ and $\lim_{h \rightarrow 0} \frac{cos(h)-1}{h} = 0$

Which can be proved in any number of ways. For example, the second of these is actually the derivative of cos at 0:

$\lim_{h \rightarrow 0} \frac{cos(h)-1}{h} = \lim_{h \rightarrow 0} \frac{cos(h)-cos(0)}{h-0} = cos'(0) = -sin(0) = 0$

That is mathematics.

(And being a computer programmer is no excuse!)

 

[RUber]

As PeroK wrote, the limit of a continuous function is always defined and equal to the value of the function at the limit point.
The limit should be written $\lim_{h\to 0} \sin h$ no if, no then. And it is said, the limit as h goes to zero of sine of h.
The left/right limits have to be evaluated for discontinuous functions like 1/x.

Can you plot the sin and coz functions? Visually, the derivatives and limits are rather easy to see.

 

[RUber]

if h = 0 then sin(h)/h is undefined
if $\lim_{h \to 0}$ then sin(h)/h = 1. This what we were taught in high school. Now, I just want to know 'why'

Anything divided by zero is undefined--always. However, if the function is continuous on either side of the singularity (division by zero), then you can still do most things you would like to do with that function by just defining a value for the function at zero.
Example $f(x) = sin x / x$ is undefined at zero, but
$g(x)=\left\{ \begin{array}{l l} 1& if\, x=0\\ (\sin x)/x&if\, x\neq 0 \end{array}\right.$
Is equal to f(x) almost everywhere and is also continuous for all x.

 

[jbriggs444]

I would like to draw attention to what PeroK pointed out in response #4 and which does not seem to have sunk in.

The notation $\lim_{h \to 0} \sin(h) = 0$ is not a statement about h. It is not a statement about the value of sin(h) for any particular h. It is not a statement about the value of sin(0). Instead, it is a statement about the values that the $\sin$ function takes on in the neighborhood of zero. It is (roughly) a statement that as the neighborhoods around 0 get very very small, the values of the sin function within such neighborhoods are all very very close to zero.

Tearing the notation in two and treating it as if it were a statement about the value of the sin function at a single value is not correct.

 

[Stephanus]

What you have written still makes no mathematical sense. For example, as sin and cos are continuous functions we have:[..]

Thanks a lot. I'll remember all those equations thet you wrote. It's not memorizing. It's the 'misleading' that I'm afraid of.. Supposed, if I consider h = 0 and 6h/h = 6., then, I'll be mislead forever.
Of course being a computer programmer is no excuse, everybody has to grasp and grab basic math. What excuse is that I wrote
"if the derivative of sin(x) is cos(x) then.." While what I meant is "Because the derivative of sin(x) is cos(x) then the proof is ..."
"IF" implies that the derivative of sin(x) is might/might not be cos(x).
Believe me, it was typo. We were taught since high school that sin(x) is cos(x), but here in PF Forum for the first time the mentors/advisors enlight me the proof!. I must have missed class and fooled around that day

 

[Stephanus]

As PeroK wrote, the limit of a continuous function is always defined and equal to the value of the function at the limit point.
The limit should be written $\lim_{h\to 0} \sin h$ no if, no then. And it is said, the limit as h goes to zero of sine of h.
The left/right limits have to be evaluated for discontinuous functions like 1/x.

Thanks a lot.

Can you plot the sin and coz functions? Visually, the derivatives and limits are rather easy to see.

Plot? What do you mean by plot? Drawing it? Well, it will be difficult with the tools in my computer.
But sin and cos curve are like wave, right?
Sine, with peak at pi/2 and trough at 0, Pi, 2Pi, 3Pi, etc...
And if we shift the cosine curve pi/2 to the left/right, it will match sine curve. I hope that is right.

 

[Stephanus]

Anything divided by zero is undefined--always. However, if the function is continuous on either side of the singularity (division by zero), then you can still do most things you would like to do with that function by just defining a value for the function at zero.
Example $f(x) = sin x / x$ is undefined at zero, but
$g(x)=\left\{ \begin{array}{l l} 1& if\, x=0\\ (\sin x)/x&if\, x\neq 0 \end{array}\right.$
Is equal to f(x) almost everywhere and is also continuous for all x.

Thanks. I'll remember the concept.

 

[Stephanus]

I would like to draw attention to what PeroK pointed out in response #4 and which does not seem to have sunk in.

The notation $\lim_{h \to 0} \sin(h) = 0$ is not a statement about h. It is not a statement about the value of sin(h) for any particular h. It is not a statement about the value of sin(0). Instead, it is a statement about the values that the $\sin$ function takes on in the neighborhood of zero. It is (roughly) a statement that as the neighborhoods around 0 get very very small, the values of the sin function within such neighborhoods are all very very close to zero.

Tearing the notation in two and treating it as if it were a statement about the value of the sin function at a single value is not correct.

Ok, I'll contemplate your answer.

 

[RUber]

What tools do you have on your computer? sin and cos are very common functions that most calculators can graph for you. There are also many online options, wolframalpha.com is useful.

Cosine shifted to the right, not left, by pi/2 will match sine.

 

[Stephanus]

What tools do you have on your computer? sin and cos are very common functions that most calculators can graph for you. There are also many online options, wolframalpha.com is useful.

Cosine shifted to the right, not left, by pi/2 will match sine.

Tools? Only Microsoft Paint and Microsoft Excel. Of course I could make a small computer program to PutPixel(X,ScreenHeight-Sin(X/100*Pi)*100), but.. that will be tedious.
"Cosine shifted to the left not match sine"? I'll try to imagine that. But I want to answer your post first.
Correction:PutPixel(X,Round(ScreenHeight-Sin(X/100*Pi)*100))

 

[Stephanus]

Cosine shifted to the right, not left, by pi/2 will match sine.

That's right cosine shift to the left by 1.5Pi.
Peat at 1 and trough at -1, right. I imagined Peat at 1 and trough only at zero. Okay the period of the graph for cosine to sine is 0.5pi to the right.
[Add: Why didn't you correct my post?

Sine, with peak at pi/2 and trough at 0, Pi, 2Pi, 3Pi, etc...

]

 

[RUber]

Excel has okay built-in plotting tools. This plot of sin(x) took about 1 minute to make.

Functions that might be useful are sin(), cos(), and pi().

 

[Mark44]

Perhaps the word "IF" is the wrong word for mathematic.
I should have written.
Because the derivative of Sin(x) is Cos(x) so the proof is:
$\lim_{h \to 0} \sin(x)\frac{\cos(h)-1}{h}$ should be zero. then, etc...

You have that backwards.
The derivative of sin(x) is cos(x), in part because of the limit above, and in part because $\lim_{h \to 0}\frac{sin(h)}{h} = 1$. In other words, the derivative of sin(x) is cos(x) because of the two limits, not the other way around.

But, I'm a computer programmer, I use if ... then ... instinctively.

Really? Even for loops or for statements that should execute one after the other? When you write code, do you always start off with "if ... then"?

Thanks a lot. I'll remember all those equations thet you wrote. It's not memorizing. It's the 'misleading' that I'm afraid of.. Supposed, if I consider h = 0 and 6h/h = 6., then, I'll be mislead forever.
Of course being a computer programmer is no excuse, everybody has to grasp and grab basic math. What excuse is that I wrote
"if the derivative of sin(x) is cos(x) then.." While what I meant is "Because the derivative of sin(x) is cos(x) then the proof is ..."
"IF" implies that the derivative of sin(x) is might/might not be cos(x).
Believe me, it was typo. We were taught since high school that sin(x) is cos(x)

Hopefully what you meant to say was "the derivative of sin(x) is cos(x)"

, but here in PF Forum for the first time the mentors/advisors enlight me the proof!. I must have missed class and fooled around that day

 

[Stephanus]

You have that backwards.
The derivative of sin(x) is cos(x), in part because of the limit above, and in part because $\lim_{h \to 0}\frac{sin(h)}{h} = 1$. In other words, the derivative of sin(x) is cos(x) because of the two limits, not the other way around.

Right! Now I think I like math language more than natural language. No ambiguity in math. Except for sinh^-1(x), that mistakenly lead me to a wrong calculation.

Really? Even for loops or for statements that should execute one after the other? When you write code, do you always start off with "if ... then"?

Of course no. But it is planted in your way of thinking. I was thinking "If the derivative of sin(x) is cos(x) then I HAVE to find the proof to show that the derivative of sin(x) is cos(x)" Not that the derivative of sin(x) might not be cos(x).Well, "backward" is the right word. Because we are already shown that the derivative of sin(x) is cos(x), but rarely the proof.

Hopefully what you meant to say was "the derivative of sin(x) is cos(x)"

Another typo, sorry . Just realize now.