[MUSIC]. Remember, continuity is all how nearby
inputs are sent to nearby outputs. Differentiability is how wiggling the
input affects the output. In light of this, they seemed related,
right? Something like the following seems
plausible. Here's the theorem.
Theorem. If f is differentiable at a, then f is
continuous at a. In other words, a differentiable function
is continuous. Morely, we know that a differentiable
function is continuous. But were advanced enough at the is point
in the course to give a precise argument using limit.
Here we go. Let's suppose that f prime of a exists.
In other words, that means a certain limit exists.
What limit? Well, the limit of f(x)-f(x)/x-a as x approaches a.
This limit of a difference quotients computes the derivative for the function
at a. So, to say that the derivative exists is to say that this limit exists.
Now, here comes the trick. What I'd like to compute is the limit of
f(x)-f(a) as x approaches a, but I don't know how to do that directly.
But, I can rewrite this thing I'm taking the limit of as a product.
Watch. Instead of taking this limit, I'm going
to take the limit as x approaches a of x-a times this difference quotient, times
f(x)-f(a)/x-a. Now, as long as x isn't equal to a, this
product is equal to this difference. Now, why does that help? Well, this is a
limit of a product. So, by one of the limit laws, the limit
of a product's the product of the limits as long as the limits exist.
And in this case, they do. So, this limit of this product is, the
product of the limits. It's the limit of x-a as x approaches a,
times the limit of f(x)-f(a)/x-a. I'm only allowed to use this limit law
because I know both of these limits exist.
Now, this first limit, the limit of x-a as x approaches a, that's 0.
And this second limit, well, this limit exists precisely because I'm assuming
differentiability, the function's are differentiable.
So, this limit is calculating the derivative at a, and zero times any
number is equal to zero. The upshot here is that we've shown that
the limit of f(x)-f(a)=0 as x approaches a.
Why would you care about this? How does that help us? We know that the limit of
f(x)-f(a) as x approaches a is equal to 0.
What that means is that the limit of f(x) as x approaches a is equal to f(a), but
this is just the definition of continuity.
So now we know that f is continuous at the point a.
That's where we ended up. Remember, what we started with.
We started by assuming that f was differentiable at a.
And after doing all this work, we ended up concluding that f is continuous at the
point a. So, differentiability implies continuity.
One way to keep track of arguments like this is to think about clouds and rain.
Theorem. If it is rainy, then it is cloudy.
A shorter way of saying this, rainy implies cloudy.
Now the question is, does it go the other way? If it's cloudy, is it necessarily
rainy? Can you think of a cloudy day with no rain? Yes, today.
Let's look out the window. It is very cloudy but there's no rain.
Beyond clouds and rain, let's bring this back to the mathematics. A differentiable
function is continuous, can you think of a continuous function which isn't
differentiable? You might want to hit pause right now,
if you don't want the puzzle given away. Here's an example of a function which is
continuous but not differential, the function f(x)=|x|.
We recently saw that the absolute value function wasn't differentable at zero.
But how do we know that the absolute value function is continuous everywhere?
We know that the absolute value function is continuous.
I mean, look at it. It's all one piece.
But, we can do better. We can use our limit knowledge to make a
more precise argument. We know that f(x)=|x| is continuous for
positive inputs. It's continuous on the open interval from
zero to infinity because the function x is continuous there.
And this function, the absolute value function, agrees with the function x if I
plug in positive numbers. Likewise, I know that the function is
continuous on negative inputs because the function -x is continuous there, and the
function -x agrees with this function on this interval.
The only sticking point is to check that the function's continuous at zero. And if
I know it's continuous for positive inputs, negative inputs, and it's
continuous at zero, then I know that it's continuous for all inputs.
Now, how do I know that the absolute value function is continuous at zero?
Well, that's another limit argument, right? The limit of the absolute value
function when I push from the right-hand side is the same as the limit of the
absolute value function when I push from the left-hand side, they're both zero.
And because these two one-sided limits exist and agree, then I know the
two-sided limit of the absolute value function is equal to zero,
which is also the function's value at zero.
And therefore, the abslute value function is continuous.
In the end, there's some relationship between differentiability and continuity.
Differentiable functions are continuous. Mathematics isn't just a sequence of
unrelated concepts. [MUSIC] It's a single unified whole.
All of these ideas are connected at the deepest possible levels.
[MUSIC]