Question

1. let ( f ) be a function such that ( lim_{h \to 0} \frac{f(2+h) - f(2)}{h} = 5 ). which of the following must be true?

i. ( f ) is continuous at ( x = 2 ).
ii. ( f ) is differentiable at ( x = 2 ).
iii. the derivative of ( f ) is continuous at ( x = 2 ).
(a) i only \t(b) ii only \t(c) i and ii only \t(d) i and iii only \t(e) ii and iii only
calculator allowed

Explanation:

Step1: Recall the definition of derivative

The limit \(\lim_{h \to 0}\frac{f(2 + h)-f(2)}{h}\) is the definition of the derivative of \(f\) at \(x = 2\), i.e., \(f^{\prime}(2)\). So if this limit exists (and it is given as 5), then \(f\) is differentiable at \(x=2\) (this is the definition of differentiability at a point). So II is true.

Step2: Recall the relationship between differentiability and continuity

A theorem in calculus states that if a function is differentiable at a point, then it is continuous at that point. Since we just established that \(f\) is differentiable at \(x = 2\) (because the derivative limit exists), then \(f\) must be continuous at \(x=2\). So I is true.

Step3: Analyze the continuity of the derivative

The existence of \(f^{\prime}(2)\) (i.e., the derivative at \(x = 2\)) does not imply that the derivative function \(f^{\prime}(x)\) is continuous at \(x = 2\). There are functions where the derivative exists at a point, but the derivative function is discontinuous at that point. For example, consider a function with a "corner" in its derivative graph or a function where the derivative has a jump discontinuity but the derivative at the point still exists. So we cannot conclude that the derivative is continuous at \(x = 2\) just from the existence of \(f^{\prime}(2)\). So III is not necessarily true.

Answer:

C. I and II only