Using Limit Definition of the Derivative?

  • #1

Main Question or Discussion Point

If one uses the limit definition of a derivative (lim of (f(x)-f(a)) / (x-a)) as x approaches a) on a function and you get a value (ie. it is not undefined) does that mean the derivative of the function at that point exists? In other words, even if the limit definition of the derivative works, do you still need to determine whether the function is continuous, smooth and non-vertical at x=a in order to know that the function is differentiable at x=a?

For example,

The derivative of f(x)=(x^2)(sinx) at x=0 is 0 (using limit definition). Is that all the proof needed to show that the function is differentiable at x=0?
 

Answers and Replies

  • #2
lavinia
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A function is differentiable at a point if the limit of the Newton quotients exist at that point.
try to convince yourself that the function is automatically continuous at the point
 
  • #3
Simon Bridge
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If one uses the limit definition of a derivative (lim of (f(x)-f(a)) / (x-a)) as x approaches a) on a function and you get a value (ie. it is not undefined) does that mean the derivative of the function at that point exists?
http://www.mathcs.org/analysis/reals/cont/derivat.html
You'd normally just say $$f^\prime(a)=\lim_{(x-a)\rightarrow 0}\frac{f(a+(x-a))-f(a)}{x-a}$$... follows from the definition of a derivative. If the limit exists then the function is differentiable at point a by definition. (I wrote it like that to draw a link with the general definition of the derivative.))

In other words, even if the limit definition of the derivative works, do you still need to determine whether the function is continuous, smooth and non-vertical at x=a in order to know that the function is differentiable at x=a?
Well, in each of those cases, the limit won't converge will it? Well... the above is basically a one-sided limit: see below.

The derivative of f(x)=(x^2)(sinx) at x=0 is 0 (using limit definition). Is that all the proof needed to show that the function is differentiable at x=0?
In this case, yep.
However, it gets conceptually hairy when we include things like the Cantor function.

Is the Cantor function "continuous"? Is it differentiable?
 
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