Differentiability using limit definition

[Qube]

Homework Statement

http://i.minus.com/jbzvT5rTWybpEZ.png

Homework Equations

If a function is differentiable, the function is continuous. The contrapositive is also true. If a function is not continuous, then it is not differentiable.

A function is differentiable when the limit definition of its derivative exists.

A limit exists when the left and right hand limits are equal.

The limit of sinx/x as x approaches 0 is 1.

The Attempt at a Solution

Okay I first tried making the function continuous but I found that the function would be continuous for 0. I didn't think this was the right answer because the question asked me to use the limit definition so I applied that. Plus, the converse of the statement if a function is diff. it is continuous isn't necessarily true.

The left hand limit is equal to 2. The limit of (2sinx - 0) / (x - 0 ) is 2, using the trig identity mentioned above.

The right hand limit only equals 2 when k = 2. The limit of kx / x as x approaches 0 is k, since the x-terms cancel, and the limit is equal to k. K therefore must equal 2 for the left hand and the right hand limits to be equal.

Is this the correct solution and approach to this proble?

 

[pasmith]

Homework Statement

http://i.minus.com/jbzvT5rTWybpEZ.png

Homework Equations

If a function is differentiable, the function is continuous. The contrapositive is also true. If a function is not continuous, then it is not differentiable.

A function is differentiable when the limit definition of its derivative exists.

A limit exists when the left and right hand limits are equal.

The limit of sinx/x as x approaches 0 is 1.

The Attempt at a Solution

Okay I first tried making the function continuous but I found that the function would be continuous for 0. I didn't think this was the right answer because the question asked me to use the limit definition so I applied that. Plus, the converse of the statement if a function is diff. it is continuous isn't necessarily true.

The left hand limit is equal to 2. The limit of (2sinx - 0) / (x - 0 ) is 2, using the trig identity mentioned above.

The right hand limit only equals 2 when k = 2. The limit of kx / x as x approaches 0 is k, since the x-terms cancel, and the limit is equal to k. K therefore must equal 2 for the left hand and the right hand limits to be equal.

Is this the correct solution and approach to this proble?

Yes.

 

[brmath]

You are correct that to make this differentiable you have to pick k = 2. And you are correct in your reasoning about how you got there.

The discussion about continuity is not relevant. You weren't asked to make it continuous or show that it is continuous. You are writing down correct facts about continuity, but perhaps a little context is needed? For a function to be differentiable is a stronger condition than for it to be continuous. Lots of functions like f(x) = |x| are continuous but not differentiable at one or several points. Someone has even defined a horror which is continuous everywhere and differentiable nowhere. So continuous cannot imply differentiable.

But differentiability is about smoothness. If lim$_{x \rightarrow x_0} \frac {f(x) - f(x_0)}{x-x_0}$ is to exist, how could f be discontinuous at $x_0$? Either the left and right limits won't be the same, or $f(x) - f(x_0)$ doesn't go to zero. (Can you see why?). Note also that no one is saying f has to be continuous anywhere near $x_0$. It could have all sorts of discontinuities nearby. But if it's differentiable at $x_0$ it is continuous at that point.

 

[HallsofIvy]

This problem specifically said "use the limit definition" but if it had not you could have used the fact that, while the derivative of a function is NOT necessarily continuous, it does satisfy the "intermediate value property". A consequence of that is that either \lim_{x\to a} f'(x)= f'(a) or the limit is does not exist- there cannot be a jump discontinuity. Here f'(x)= 2cos(x) if x is less than 0, k if x is greater than 0. The limit from below is 2 so the limit from above must also be 2: k= 2.

 

[brmath]

This problem specifically said "use the limit definition" but if it had not you could have used the fact that, while the derivative of a function is NOT necessarily continuous, it does satisfy the "intermediate value property". A consequence of that is that either \lim_{x\to a} f'(x)= f'(a) or the limit is does not exist- there cannot be a jump discontinuity. Here f'(x)= 2cos(x) if x is less than 0, k if x is greater than 0. The limit from below is 2 so the limit from above must also be 2: k= 2.

Proving that derivatives satisfy the intermediate value property is not easy, and surely beyond the scope of this course. I remember pouncing on that theorem when I first saw it, sure that it would be very useful. And no doubt it is of use somewhere. But actually I myself have never used it for anything -- there was always and easier and cleaner way to do whatever.