Proof limit derivatave of sin(x)

[Stephanus]

sin(0) = 0 so we do not need to consider limits.

I mean $\lim_{h \to 0} \sin(h)$

Is there a limit in $\lim_{h \to 0} \sin(h)$?
What is the result? 0 or ≈ 0?
Left/Right hand is not equal, right?

cos(0) = 1 so we do not need to consider limits.

I mean $\lim_{h \to 0} \cos(h)$

Is there a limit in $\lim_{h \to 0} \cos(h)$?
Left/Right hand is equal, right.
The result? 1 or ≈ 1?

sin(0) = 0 so we do not need to consider limits.

Yes, but because $(x + h) - x = (h + x) - x = h + (x - x) = h$ we would write $$ \frac{d(\sin x)}{dx} = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h} $$[..][/QUOTE]Still reading the rest...

 

[micromass]

Is there a limit in $\lim_{h \to 0} \sin(h)$?
What is the result? 0 or ≈ 0?

It is =0.

Left/Right hand is not equal, right?

They are equal.

Is there a limit in $\lim_{h \to 0} \cos(h)$?

=1

Left/Right hand is equal, right.

Yes.

 

[pbuk]

I mean $\lim_{h \to 0} \sin(h)$

I know you do. Because sin(h) is continuous, in particular it is continuous at h = 0 (which implies that it is defined at h = 0), by definition $\lim_{h \to 0} \sin(h) = \sin(0) = 0$, so we don't need to do any working out we simply write sin(0) = 0. Likewise by definition $\lim_{h \to 0} \cos(0) = \cos(0) = 1$.
We can't do this with $\lim_{x \to 0} \frac{\sin x}{x}$ because $\frac{\sin x}{x}$ is not defined at x = 0.

(Edited to improve accuracy of statements regarding definition/continuity)

 

[Stephanus]

I know you do. Because sin(h) is continuous, in particular it is continuous at h = 0, by definition $\lim_{h \to 0} \sin(h) = \sin(0) = 0$, so we don't need to do any working out we simply write sin(0) = 0. Likewise by definition $\lim_{h \to 0} \cos(0) = \cos(0) = 1$.
We can't do this with $\lim_{x \to 0} \frac{\sin x}{x}$ because $\frac{\sin x}{x}$ has a discontinuity at x = 0.

So, "continuous", "left hand", "right hand". Ok, I'll study them.
Btw, tangent/cotangent are not continuous, right?

 

[micromass]

So, "continuous", "left hand", "right hand". Ok, I'll study them.
Btw, tangent/cotangent are not continuous, right?

They are continuous wherever they are defined.

 

[Stephanus]

Who "they"? Tangent?
But Tangent 0 is undefined, right.

 

[micromass]

Who "they"? Tangent?

They = tangent and cotangent.

But Tangent 0 is undefined, right.

The tangent in 0 is not undefined.

 

[Stephanus]

Who "they"? Tangent?
But Tangent 0 is undefined, right.

You mean tangent pi/2 - is ∞ and tangent pi/2+ is ∞ so it's said that they are continuous. It's tangent pi/2 is undefined, right. Sorry, to hasty to type.

 

[micromass]

You mean tangent pi/2 - is ∞ and tangent pi/2+ is ∞

I have no idea what that means. $\tan(\pi/2)$ is undefined. The limits are
\lim_{x\rightarrow \pi/2 - } \tan(x) = +\infty~\text{and}~\lim_{x\rightarrow \pi/2+} \tan(x) = -\infty

 

[Mark44]

You mean tangent pi/2 - is ∞ and tangent pi/2+ is ∞

I have no idea what that means. $\tan(\pi/2)$ is undefined. The limits are
\lim_{x\rightarrow \pi/2 - } \tan(x) = +\infty~\text{and}~\lim_{x\rightarrow \pi/2+} \tan(x) = -\infty

And by the two limits just above, micromass means the limit as x approaches $\pi/2$ from the left in the first limit, and the limit as x approaches $\pi/2$ from the right, respectively. Since these limits are as different as they could possibly be, the two-sided limit does not exist.

 

[Stephanus]

And by the two limits just above, micromass means the limit as x approaches $\pi/2$ from the left in the first limit, and the limit as x approaches $\pi/2$ from the right, respectively. Since these limits are as different as they could possibly be, the two-sided limit does not exist.

Yes you're right.

 

[Stephanus]

I mean $\lim_{h \to 0} \sin(h)$

It is =0. They are equal.

Yes. I just realize now. Sin (-h) is negative and Sin(+h) is positive.
Sin(-x) = -Sin(x), while
Cos(x) = Cos(-x). Shouldn't forget that.

 

[Mark44]

Sin (-h) is negative and Sin(+h) is positive.

You really need to specify that h is near zero and positive in the above statements. For example, $\sin(-3\pi/2) > 0$ and $\sin(3\pi/2) < 0$.

When you are using variables, attaching - to a variable doesn't make it negative any more than attaching + to it makes a positive number.

 

[Stephanus]

You really need to specify that h is near zero and positive in the above statements. For example, $\sin(-3\pi/2) > 0$ and $\sin(3\pi/2) < 0$.

When you are using variables, attaching - to a variable doesn't make it negative any more than attaching + to it makes a positive number.

You are right! h is lim h -> 0

 

[Stephanus]

Hmm, very insightful post. To me it seemed straight forward because I can "feel" the geometry so to speak. Yes, I can understand your point of how geometry is not put in rigorous footing in these classes.

So, what would be an adequate proof of lim sinx/x? Can you inbox me. I feel it will derail the intention of this thread.

No, not Inbox. Here is okay. I'll read it anyway even if I don't understand it at once. I just read it in my spare time.

 

[Stephanus]

As long as $\theta$ is different from zero, $\frac{\sin\theta}{\theta}$ will be different from 1. What we can say, though, is $\lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$. IOW, the limit of this fraction is 1, even though $\frac{\sin \theta}{\theta}$ is never equal to 1 for any nonzero value of $\theta$.

So,
if $\theta \neq 0$ then $\frac{\sin(\theta)}{\theta} \neq 1$
if $\theta = 0$ then $\frac{\sin(\theta)}{\theta}$ is undefined
But $\lim_{\theta \to 0}$ then $\frac{\sin\theta}{\theta} = 1$
Is that the concept of limit?
$\lim_{h \to 0}$ h is not zero, but has is not non-zero either?

 

[Mark44]

As long as $\theta$ is different from zero, $\frac{\sin\theta}{\theta}$ will be different from 1. What we can say, though, is $\lim_{\theta \to 0}\frac{\sin \theta}{\theta} = 1$. IOW, the limit of this fraction is 1, even though $\frac{\sin \theta}{\theta}$ is never equal to 1 for any nonzero value of $\theta$.

So,
if $\theta \neq 0$ then $\frac{\sin(\theta)}{\theta} \neq 1$
if $\theta = 0$ then $\frac{\sin(\theta)}{\theta}$ is undefined
But $\lim_{\theta \to 0}$ then $\frac{\sin\theta}{\theta} = 1$

Please stop inserting "if" and "then" in the middle of things where they don't belong. The limit part ($\lim_{\theta \to \text{whatever}}$) should not be separated from the thing you're taking the limit of.
Corrected, the above should read:
##\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta} = 1

Is that the concept of limit?

No, that's not the limit concept. It's an example of one specific limit.

$\lim_{h \to 0}$ h is not zero, but has is not non-zero either?

Clearly you are not getting it. When the function you're taking the limit of is a polynomial, just evaluate the polynomial at the limit number.

$\lim_{h \to 5} h = 5$
$\lim_{h \to 5} h^2 = 25$
$\lim_{h \to 3} h^3 - 1 = 26$

The limit of a constant is that constant.
$\lim_{h \to 5} 3 = 3$

Where limits come into their own is with rational functions, at points where the denominator is zero.(Limits are also important in evaluating the behavior of other types of functions.)
$\lim_{h \to 1}\frac{h^2 - 1}{h - 1} = 2$
For this example, f(h) = $\frac{h^2 - 1}{h - 1}$, which is undefined at h = 1. You can't simply plug h = 1 into the function, because you get 0/0, an indeterminate form. By the use of limits, it can be shown that a limit exists, even though the function is not defined at h = 1.
$\lim_{h \to 1}\frac{h^2 - 1}{h - 1} = \lim_{h \to 1}\frac{(h + 1)(h - 1)}{h - 1} = \lim_{h \to 1} h +1 \cdot \lim_{h \to 1}\frac{h - 1}{h - 1}$
The first limit just above is clearly 2. In the second limit, because h - 1 over itself is always 1, the limit of (h - 1)/(h - 1) is 1, even though we can't evaluate this rational function directly at h = 1.

Since the first limit in this example is 2 and the second limit is 1, the limit of the product in this example is 2.

 

[Stephanus]

Please stop inserting "if" and "then" in the middle of things where they don't belong. The limit part ($\lim_{\theta \to \text{whatever}}$) should not be separated from the thing you're taking the limit of.
Corrected, the above should read:
$\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta} = 1$

Okay, okay.
So number 1 and 2 is okay
1. If h ≠ 0 then sin(h) ≠ 0
2. If h = 0 then sin(h)/h is undefined
3. NO IF THEN, but $\lim_{\theta \to 0}\frac{\sin(\theta)}{\theta} = 1$

The limit part ($\lim_{\theta \to \text{whatever}}$) should not be separated from the thing you're taking the limit of.

I WILL REMEMBER THAT.

No, that's not the limit concept. It's an example of one specific limit.

Ok

Clearly you are not getting it. When the function you're taking the limit of is a polynomial, just evaluate the polynomial at the limit number.

$\lim_{h \to 5} h = 5$
$\lim_{h \to 5} h^2 = 25$
$\lim_{h \to 3} h^3 - 1 = 26$

The limit of a constant is that constant.
$\lim_{h \to 5} 3 = 3$

Where limits come into their own is with rational functions, at points where the denominator is zero.(Limits are also important in evaluating the behavior of other types of functions.)
$\lim_{h \to 1}\frac{h^2 - 1}{h - 1} = 2$
For this example, f(h) = $\frac{h^2 - 1}{h - 1}$, which is undefined at h = 1. You can't simply plug h = 1 into the function, because you get 0/0, an indeterminate form. By the use of limits, it can be shown that a limit exists, even though the function is not defined at h = 1.
$\lim_{h \to 1}\frac{h^2 - 1}{h - 1} = \lim_{h \to 1}\frac{(h + 1)(h - 1)}{h - 1} = \lim_{h \to 1} h +1 \cdot \lim_{h \to 1}\frac{h - 1}{h - 1}$
The first limit just above is clearly 2. In the second limit, because h - 1 over itself is always 1, the limit of (h - 1)/(h - 1) is 1, even though we can't evaluate this rational function directly at h = 1.

Since the first limit in this example is 2 and the second limit is 1, the limit of the product in this example is 2.

Thank you very much Mark44. You have given me an invaluable lesson about limit at the above quote.
Franky in high school (30 years ago?) I knew that $\lim_{h \to 0} \frac{6h}{h} = 6$, but I think I haven't BEEN taught that $\lim_{h \to 4} h = 4$. I must have gone to the wrong school.[Add: And I have a nasty habit to type faster than I think.
I have edited my post three times. After I read again.
1. $\lim_{h \to 4} is 4$ to $\lim_{h \to 4} h = 4$
because

The limit part ($\lim_{\theta \to \text{whatever}}$) should not be separated from the thing you're taking the limit of.

2. Invaluable leson -> Invaluable lesson
3. I haven't taught -> I haven't been taught
Afraid that your correction comes first and pointed me the error that I typed before I get the chance to realize it]