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### Course: AP®︎/College Calculus AB>Unit 2

Lesson 4: Connecting differentiability and continuity: determining when derivatives do and do not exist

# Proof: Differentiability implies continuity

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.
Proof: Differentiability implies continuitySee video transcript

## Want to join the conversation?

• I'm just wondering, at why did Sal choose the lim x->c(f(x)-f(c))? Was it just an arbitrary limit that he chose?
Thanks
• He chose it because someone else worked out that this procedure gave the desired proof ...

Doing proofs involves a lot of trial and error!
• 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)
• This comment is gold. Thank you
(1 vote)
• i love you Sal, ur the best for real
• At Why 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 ?
• "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)".
• 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?
• This proves that differentiability implies continuity when we look at the equation Sal arrives to at . 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.
• 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.
• 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)
• 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?
• 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.
• By using "implies" does this mean that where a function is differentiable the function must be continuous or does that mean that it strongly indicates that the function is also continuous? (assuming the must)