1. Provide an example of a function f that is continuous at a = 2 but not differentiable at a = 2.
2. Provide an example of a function g that is differentiable at a = 3 but does not have a limit at a = 3.
Thanks!
for the 2nd part, if a function is differentiable, it is continuous. Hence it has a limit at every point.
f(x) = |x-2|
1. An example of a function f that is continuous at a = 2 but not differentiable at a = 2 is:
f(x) = |x - 2|
This function represents the absolute value of the difference between x and 2. At x = 2, f(x) is equal to 0, and it is continuous at this point. However, the function is not differentiable at x = 2 because the left and right-sided derivatives do not agree. The left derivative is -1 and the right derivative is 1.
2. An example of a function g that is differentiable at a = 3 but does not have a limit at a = 3 is:
g(x) = sin(1 / (x - 3))
This function represents the sine of 1 divided by the difference between x and 3. As x approaches 3, the denominator approaches 0, and the sine function oscillates infinitely between -1 and 1. Therefore, the limit as x approaches 3 does not exist. However, the derivative of g(x) can be computed using the chain rule, and it exists for all x ≠ 3. Thus, g(x) is differentiable at x = 3.
1. To find a function that is continuous at a = 2 but not differentiable at a = 2, we need a function that satisfies the definition of continuity and fails the differentiability test at that point.
One example of such a function is f(x) = |x - 2|. The absolute value function is continuous over its entire domain. At x = 2, the function f(x) = |x - 2| gives us f(2) = |2 - 2| = 0. So, f(x) is continuous at a = 2.
To check differentiability at a = 2, we need to calculate the derivative of f(x) at x = 2. We can approach this by calculating the left and right-hand derivatives separately.
The left-hand derivative, denoted f'(2-), is the limit as x approaches 2 from the left (x < 2), and it can be calculated as:
f'(2-) = lim x->2- (f(x) - f(2))/(x - 2)
= lim x->2- (|x - 2| - 0)/(x - 2)
= lim x->2- (x - 2)/(x - 2)
= lim x->2- 1
= 1
The right-hand derivative, f'(2+), is the limit as x approaches 2 from the right (x > 2), and can be calculated as:
f'(2+) = lim x->2+ (f(x) - f(2))/(x - 2)
= lim x->2+ (|x - 2| - 0)/(x - 2)
= lim x->2+ (-(x - 2))/(x - 2)
= lim x->2+ -1
= -1
Since f'(2-) does not equal f'(2+), the function f(x) = |x - 2| is not differentiable at x = 2.
2. To find a function that is differentiable at a = 3 but does not have a limit at a = 3, we need a function that satisfies the differentiability criterion but fails the limit criterion.
One example of such a function is g(x) = 1/x. This function is differentiable everywhere except at x = 0. However, at x = 3, the function g(x) does not have a limit, as the function approaches infinity as x approaches 3 from the positive side (x > 3) and approaches negative infinity as x approaches 3 from the negative side (x < 3).
To verify that g(x) is differentiable at x = 3, we can calculate its derivative at that point. Using the quotient rule, we have:
g'(x) = (-1/x^2) * (1) - (1)(-2x)/(x^2)^2
= -1/x^2 + 2x/x^4
= (2x - 1)/x^4
By calculating the limit of g'(x) as x approaches 3, we can determine if the derivative exists:
lim x->3 (2x - 1)/x^4 = (2(3) - 1)/(3)^4 = 5/81
Since the limit of g'(x) as x approaches 3 exists and is finite, the function g(x) = 1/x is differentiable at x = 3. But it does not have a limit at x = 3.