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## Differential Calculus

### Course: Differential Calculus > Unit 2

Lesson 5: Differentiability- Differentiability and continuity
- Differentiability at a point: graphical
- Differentiability at a point: graphical
- Differentiability at a point: algebraic (function is differentiable)
- Differentiability at a point: algebraic (function isn't differentiable)
- Differentiability at a point: algebraic
- Proof: Differentiability implies continuity

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# Proof: Differentiability implies continuity

AP.CALC:

FUN‑2 (EU)

, FUN‑2.A (LO)

, FUN‑2.A.1 (EK)

, FUN‑2.A.2 (EK)

If a function is differentiable then it's also continuous. This property is very useful when working with functions, because if we know that a function is differentiable, we immediately know that it's also continuous.

The AP Calculus course doesn't require knowing the proof of this fact, but we believe that as long as a proof is accessible, there's always something to learn from it. In general, it's always good to require some kind of proof or justification for the theorems you learn.

## Want to join the conversation?

- I'm just wondering, at7:08why did Sal choose the lim x->c(f(x)-f(c))? Was it just an arbitrary limit that he chose?

Thanks(25 votes)- He chose it because someone else worked out that this procedure gave the desired proof ...

Doing proofs involves a lot of trial and error!(22 votes)

- At10:11Why did we assume that
`lim x -> c f(c) = f(c)`

?

What if function f(x) is undefined at c or if`lim x -> c f(c)`

does not exist ?(1 vote)- "What if function f(x) is undefined at c?": We assumed that f(x) is differentiable at x=c. This alone means f(x) cannot be undefined at x=c.

And as tyersome answered, f(c) doesn't have an x, so lim x->c can be removed from "lim x -> c f(c)".(10 votes)

- Arguably this is the same proof formatted in a more intuitive way:

(1) lim{x->c}{(f(x)-f(c))/(x-c)} = f'(c)

(2) lim{x->c}{(f(x)-f(c))/(x-c)} * lim{x->c}{x-c} = f'(c) * lim{x->c}{x-c}

(3) lim{x->c}{x-c} = 0

From (2) and (3) we have

(4) lim{x->c}{(f(x)-f(c))/(x-c)} * lim{x->c}{x-c} = f'(c) * 0

From (4) and the rule about limit multiplication we have

(5) lim{x->c}{((f(x)-f(c))/(x-c)) * (x-c)} = 0

(6) lim{x->c}{f(x)-f(c)} = 0

(7) lim{x->c}{f(x)} - lim{x->c}{f(c)} = 0

(8) lim{x->c}{f(x)} - f(c) = 0

(9) lim{x->c}{f(x)} = f(c)(8 votes)- i also may vote for this direction than the one used in video

by the assumption of a functions' differentiability we can start with (1) then drive toward (9)

but in video, Sal dives into the proof from (6) then goes to both directions, up and down, to get the "intended" match

but we can't talk about (6) only with its differentiability. in fact, (6) is what we're going to prove with (1). thus it feels like a circular reasoning

the core concepts used in both proofs are the same [lim(A*B) = lim(A)*lim(B) and lim(A+B) = lim(A)+lim(B)]. the difference is in flows. and i bet this one is right(0 votes)

- HELP! I've been thinking about this topic for a little bit. I've watched the video a couple of times and consulted a few other sources, and I can say that I am thoroughly confused.

How exactly does this proof show that Differentiability implies continuity? I can follow the math. But, I think that I'm missing something here. What are the key parts of this proof and how does it link together the topics of differentiability and continuity?(2 votes)- This proves that differentiability implies continuity when we look at the equation Sal arrives to at8:11. If the derivative does not exist, then you end up multiplying 0 by some undefined, which is nonsensical. If the derivative
**does**exist though, we end up multiplying a 0 by f'(c), which allows us to carry on with the proof.(12 votes)

- Just curious to know how others interpret ?

Intuitively why limit x -> c (f(x) - f(c)) = 0. I understand the math but want to know what does that mean, and why it becomes 0 intuitively ? Thanks.(2 votes)- You're basically subtracting one number from itself the limit of f(x) as x -> c means eventually f(x) will equal f(c). so that means the subtraction problem becomes f(c) - f(c)(5 votes)

- I am trying to understand what “Differentiability Implies Continuity” means. I made the incorrect assumption that to prove differentiability, all I needed to do was confirm that:

lim x->c- (f(x) – f(c))/(x – c) = lim x->c+ (f(x) – f(c))/(x – c)

The assumption was incorrect. All the above indicates is that the slope as ‘x’ approaches ‘c’ from the left and right is the same. It does not indicate that the function is continuous at ‘c’, i.e. lim x->c f(x) = f(c). It seems that to prove differentiability, it is necessary to first prove continuity; which seems counter to the statement that “Differentiability Implies Continuity”. What am I missing?(0 votes)- I think your assumption, "lim x->c- (f(x) – f(c))/(x – c) = lim x->c+ (f(x) – f(c))/(x – c)" implies that "continuity implies differentiability". However, Khan showed examples of how there are continuous functions which have points that are not differentiable. For example, f(x)=absolute value(x) is continuous at the point x=0 but it is NOT differentiable there. In addition, a function is NOT differentiable if the function is NOT continuous. In this video, Khan is merely proving that if you know the function is differentiable, then it MUST also be continuous for all the points at which it is differentiable.(8 votes)

- at2:51is the sentence "the function f(x) is differentiable at x=c... really just saying this llimit right over here actually exists" is it indicates that if the function f(x) is differentiable at x=c, then the limit exists and if the limit exists, then the function f(x) is differentiable at x=c, which is known as biconditional statement(iff)?(1 vote)
- It is not bi-directional,

In case if f(x) is differentiable at x=c, then limit exist, yes and and f(x) is continuous at x(proved in the above theorom).

And in case if f(x) is said to be continuous, we can't simply say it is differentiable, since its not just limit of f(x) we are calculating but limit of slope of f(x) is what we need to calculate to find differentiability property of f(x).

Hope this helpful.(1 vote)

- is f(c) just a point in the graph?do we define limits also for a point?(2 votes)
- yes, f(c) is the y value of the point ( c, f(c) ).(1 vote)

- @3:24definition of continuity : lim x->c f(x) = f(c)

If this holds; lim x->c f(x) - f(c) = 0, then shouldnt the numerator of differentiation quotient always be zero at point c?

lim x->c f(x) - f(c) / x-c

So then the derivative would be zero at that point algebraicaly but we know it doesnt have to be? Please comment what I'm interpreting wrong here ?(1 vote)- The differentiation limit, lim x->c (f(x)-f(c))/(x-c), will always be 0/0. However, 0/0 is indeterminate and can equal all real numbers as well as +/- infinity. What 0/0 equals depends on the specific function we are differentiating.(3 votes)

- Hi, I got lost at7:10. why did sal multiply f-c ?(1 vote)
- he multiplied and divided by f-c. THIS is because if you multiply and divide by the same number it is the same as multiplying by 1, so it doesn't change the value. for instance 2 = 2 * (3/3) = 6/3. or with variables 2x * (x/x) = 2x^2 / x

The reason he chose f-c specifically is because he was able to get limit of [f(x) - f(c)]/(x-c) as x -> c which is the derivative of f(x), which is what he wanted.

Does that help?(2 votes)