Problem Set 1: 1D-10
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Question

Problem Set 1: 1D-10
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anonymous · Answer

$$g(h)=\left[ f(a+h)-f(a) \right]/h$$has a removable discontinuity at h=0
$$f \prime(a) exist$$

anonymous · Answer

I gave it a try, is anyone convinced?
$$g(0^{-})=\lim_{h \rightarrow 0}\left[ f(a+h)-f(a) \right]/h=f \prime (a)$$
$$g(0^{+})=\lim_{h \rightarrow 0}\left[ f(a+h)-f(a) \right]/h=f \prime (a)$$
Since the left-hand limit=right-hand limit at h=0, i.e.
$$f \prime(a)=f \prime(a) $$
Therefore f'(a) exists and g(h) has a removable discontinuity at h=0

Sounds about right? seems too simple an answer...

anonymous · Answer

Yes, it is correct. The first problem set goes over some basics of high school calculus since some of the students have not taken calculus yet. So keep on going if you're fairly confident with your answer!

stacey · Answer

What you have is correct. It is removable because by defining g(0) = f'(a) the function becomes continuous at h=0.

anonymous · Answer

Thanks a million! I'm using these videos to prepare for A-levels so I'm bound to be out of my depth most of the time.