Question

Problem Set 1, Question 1D-10*

Show that

Answer 1

\[g(h) = \frac{ f(a+h) - f(a) }{ h } \] has a removable discontinuity at h=0 <=> f'(a) exists.

Can someone please explain what the question means? I understand what a removable discontinuity is. How should I approach this problem?

Thank you.

Answer 2

Here is a restatement of the question:

Let g(h) be the following function:\[g(h)=\frac{ f(a+h)-f(a) }{ h }\]Show that if g(h) has a removable discontinuity at h=0, then f'(a) exists, and also that if f'(a) exists, then g(h) has a removable discontinuity at h=0. (The double arrow here means either one implies the other.)

Answer 3

so would I be right in thinking that since the definition of removable discontinuity is that the limit of the left and right as h goes to 0 is the same and THAT is the same definition as the differentiation function g(h)?

and since f'(a) exists, the limit of f'(a) = lim h -> 0 of g(h). We also know that g(h) is discontinuous at h=0 but since the limit from the left and right are the same as h->0, we can say it's removable discontinuity?

Answer 4

You have the concept, but your statement of the first part is perhaps less clear than it might be. I would begin by saying that g(h) is defined in such a way as to be a difference quotient for f(a), with h serving as the difference in a (that is, delta a). Because this expression is in the form of a difference quotient for f(a), we can conclude that f'(a) exists if the limit as h approaches zero exists. (And so on.)

Answer 5

Yes, that does make it clearer lol. Thanks again, Creeksider!