Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

Problem Set 1: 1D-10 Show that

OCW Scholar - Single Variable Calculus
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
\[g(h)=\left[ f(a+h)-f(a) \right]/h\]has a removable discontinuity at h=0 \[f \prime(a) exist\]
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...
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!

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question