The function f(x) = x^2 sin(1/x) for x ≠ 0 and f(0) = 0 is differentiable at x =

₹0.0

Question: The function f(x) = x^2 sin(1/x) for x ≠ 0 and f(0) = 0 is differentiable at x = 0. True or False?

Options:

Correct Answer: True

Solution:

True, as the limit of f\'(x) as x approaches 0 exists and equals 0.

The function f(x) = x^2 sin(1/x) for x ≠ 0 and f(0) = 0 is differentiable at x = 0. True or False?

The function f(x) = x^2 sin(1/x) for x ≠ 0 and f(0) = 0 is differentiable at x = 0. True or False?

Step 1: Understand the function f(x) = x^2 sin(1/x) for x ≠ 0 and f(0) = 0.
Step 2: To check if f(x) is differentiable at x = 0, we need to find the derivative f'(0).
Step 3: The derivative f'(0) is defined as the limit of (f(x) - f(0)) / (x - 0) as x approaches 0.
Step 4: Since f(0) = 0, we can rewrite the limit as limit of f(x) / x as x approaches 0.
Step 5: Substitute f(x) into the limit: limit of (x^2 sin(1/x)) / x as x approaches 0.
Step 6: Simplify the expression: limit of x sin(1/x) as x approaches 0.
Step 7: As x approaches 0, sin(1/x) oscillates between -1 and 1, so x sin(1/x) approaches 0.
Step 8: Therefore, the limit exists and equals 0, which means f'(0) = 0.
Step 9: Since the derivative exists at x = 0, f(x) is differentiable at that point.

Differentiability at a Point – The question tests the understanding of whether a function is differentiable at a specific point, in this case, x = 0.
Limit of the Derivative – It requires knowledge of how to compute the derivative of a function using limits, particularly at points where the function is defined piecewise.
Behavior of Functions Near Singularities – The function involves a term sin(1/x), which can be tricky as x approaches 0, testing the ability to analyze oscillatory behavior.