# Help understanding MIT Open Course Ware Video Content

1. Dec 26, 2011

### calcstudent1

In the below video from approx. 12:00 onwards while explaining the proof for Quotient Rule, Mr. Miller says that while delta x approaches zero, delta v approaches zero. Can someone explain me why is that?

Thanks a lot!

Last edited by a moderator: Sep 25, 2014
2. Dec 26, 2011

### Fredrik

Staff Emeritus
Because $\Delta v=v(x+\Delta x)-v(x)$, and by definition of "continuous", if v is continuous, then $v(x+\Delta x)\to v(x)$ as $\Delta x\to 0$. Note that we don't have to assume that v is continuous, since this is implied by the assumption that v is differentiable.

3. Dec 26, 2011

### calcstudent1

I understand that delta v = new v minus old v but the thing I dont understand is "if v is continuous then v(x+Δx)→v(x) as Δx→0." can you explain this statement. I understand that as delta x approaches zero, delta new v is same as old v but what does he bring continuous into picture?

4. Dec 26, 2011

### Edgardo

The question is: Why is "v(x+Δx)→v(x) as Δx→0" valid?
In other words, why is the following correct: $\text{lim}_{x \rightarrow a} v(x + \Delta x) = v(x)$

For this to be valid you need v(x) to be a continuous function.
A function is continuous if you can pull the limit into the argument:
$\text{lim}_{x \rightarrow a} f(x) = f(\text{lim}_{x\rightarrow a} x)$

Let's apply this to v(x):
$\text{lim}_{\Delta x \rightarrow 0} v(x + \Delta x) = v(\text{lim}_{\Delta x\rightarrow 0} (x+\Delta x)) = v(x)$,

where $\text{lim}_{\Delta x\rightarrow 0} (x+\Delta x) = x$

5. Dec 26, 2011

### calcstudent1

wow. this sounds way too complicated for me. i am now feeling that i know so little about limits :(

can you please explain me in simple terms?

6. Dec 26, 2011

### chiro

It might be helpful for you to draw a few continuous curves to convince you that if you look at some point and then go from the left and the right of that point, that as you move closer and closer to the point, that it converges to the point no matter if you approach it from the left or the right.

If something is differentiable then its a lot easier to do this intuitively. If it is not, then you can get potentially some really complicated functions that are harder to deal with.

The intuitiveness can help you understand the actual limit laws because when you need to prove it, you will need to use these laws since you can't rely on drawing nice graphs for general purpose problems.

As a counterexample for seeing where limits don't work, draw a few graphs that are discontinuous and you will see how limits break down for these cases (the left and right hand limits are not equal).

7. Dec 26, 2011

### Edgardo

This topic is not that easy if you want to learn it the rigorous way. So, don't worry too much if you don't get it immediately .

I'll go a little back. You wrote:
The MIT professor uses the shorthand notation $\Delta u$ and $\Delta v$, where
$\Delta u =u(x+\Delta x) - u(x)$ and
$\Delta v =v(x+\Delta x) - v(x)$
see 7m35s.

He then lets $\Delta x$ approach zero. Intuitively this means that $u(x+\Delta x)$ will approach $u(x)$. Hence, $\Delta u$ approaches zero. The same goes for $\Delta v$.

More formally we can write:

$\text{lim}_{\Delta x \rightarrow 0}\Delta u= \text{lim}_{\Delta x \rightarrow 0}[u(x+\Delta x)-u(x)]$ = 0

Read this as: The limit of "u(x+Δx)-u(x)" as Δx approaches zero equals zero.

But I'll ask you this question: Why should $u(x+\Delta x)$ approach $u(x)$? Or in other words, why is $\text{lim}_{\Delta x \rightarrow 0}[u(x+\Delta x)-u(x)] = 0$ valid?

This is our goal: We want to show that $\text{lim}_{\Delta x \rightarrow 0}[u(x+\Delta x)-u(x)]$ = 0 (Goal-Equation)

-----

Let's examine the expression $\text{lim}_{\Delta x \rightarrow 0}[u(x+\Delta x)-u(x)]$ carefully. We wish this expression to equal zero.

It turns out that you can turn it into:
$\text{lim}_{\Delta x \rightarrow 0}[u(x+\Delta x)-u(x)] = \text{lim}_{\Delta x \rightarrow 0}u(x+\Delta x) - \text{lim}_{\Delta x \rightarrow 0} u(x)$ (Limit-Equation)

Do you see what I've done? I have applied the limit to each addend.

Now, lets plug the (Limit-Equation) into the (Goal-Equation):
$\text{lim}_{\Delta x \rightarrow 0}u(x+\Delta x) - \text{lim}_{\Delta x \rightarrow 0} u(x)$ = 0

Let's manipulate it further:
$\text{lim}_{\Delta x \rightarrow 0}u(x+\Delta x) = \text{lim}_{\Delta x \rightarrow 0} u(x) = u(x)$

(Make sure you understand why the last equal sign is valid)

So the final form is:
$\text{lim}_{\Delta x \rightarrow 0}u(x+\Delta x) =u(x)$ (Goal-Equation 2)

However, how can you show that Goal-Equation 2 is true?

-------

We will now show that Goal-Equation 2 is true:
We assume that we can pull the limit into the argument of u. What I mean is the following:

Consider the left hand side of Goal-Equation 2: $\text{lim}_{\Delta x \rightarrow 0}u(x+\Delta x)$

Now, let's pull the limit into the argument of u, i.e. we throw the limit between the brackets:
$\text{lim}_{\Delta x \rightarrow 0}u(x+\Delta x) = u( \text{lim}_{\Delta x \rightarrow 0} [x+\Delta x])$

Assumption: We are allowed to pull the limit into the argument

Since $\text{lim}_{\Delta x \rightarrow 0} [x+\Delta x] = x$, we can write:
$\text{lim}_{\Delta x \rightarrow 0}u(x+\Delta x) = u( \text{lim}_{\Delta x \rightarrow 0} [x+\Delta x]) = u(x)$.

Thus, we have arrived at our Goal-Equation 2.
Note however: We have used an assumption. This assumption is also called continuity of a function.

Prof. Francis Su mentions this in his excellent lecture on Real Analysis at 48m07s. This lecture involves rigorous mathematics, i.e. it is very formal and uses proofs.

8. Dec 26, 2011

### Fredrik

Staff Emeritus
You have already received a few answers, so I'll just show you an example of what can happen when v isn't continuous. Suppose e.g. that v is the function defined by $$v(x)= \begin{cases} 1 &\text{if }x=0\\ 0 &\text{if }x\neq 0. \end{cases}$$ This function is continuous at all points in its domain except 0. We have $v(0)=1$, but $\lim_{\Delta x\to 0} v(0+\Delta x)=0$.

9. Dec 27, 2011

### calcstudent1

+1.

You are awesome. Now, I need to sit and digest this concept. I understand bit and pieces but now I need to put everything together.

10. Dec 27, 2011

### calcstudent1

+1

Thanks for clarification.

11. Dec 27, 2011

### calcstudent1

You guys rock!!

12. Dec 27, 2011

### Fredrik

Staff Emeritus
By the way, the quotient rule can also be proved from the product rule, the chain rule and the result that d/dx(1/x)=-1/x2.
$$\left(\frac{f}{g}\right)' =\left(f\frac{1}{g}\right)' =f'\left(\frac{1}{g}\right) +f\left(\frac{1}{g}\right)' =\frac{f'}{g}+f\left(-\frac{1}{g^2}g'\right) =\frac{f'g-fg'}{g^2}$$

13. Dec 28, 2011

### Devil Doc

This is a great place to learn mathematics. :)