If f(x) = x^2 sin(1/x) for x ≠ 0 and f(0) = 0, is f differentiable at x = 0?
Practice Questions
1 question
Q1
If f(x) = x^2 sin(1/x) for x ≠ 0 and f(0) = 0, is f differentiable at x = 0?
Yes
No
Only left differentiable
Only right differentiable
Using the limit definition of the derivative, f'(0) exists, hence f is differentiable at x = 0.
Questions & Step-by-step Solutions
1 item
Q
Q: If f(x) = x^2 sin(1/x) for x ≠ 0 and f(0) = 0, is f differentiable at x = 0?
Solution: Using the limit definition of the derivative, f'(0) exists, hence f is differentiable at x = 0.
Steps: 9
Step 1: Understand the function f(x). It is defined as f(x) = x^2 sin(1/x) for x not equal to 0, and f(0) = 0.
Step 2: Recall the definition of differentiability at a point. A function f is differentiable at x = 0 if the limit of (f(x) - f(0)) / (x - 0) as x approaches 0 exists.
Step 3: Substitute f(0) into the limit definition. We need to find the limit of (f(x) - 0) / x as x approaches 0, which simplifies to f(x) / x.
Step 4: Rewrite f(x) for x not equal to 0. This gives us (x^2 sin(1/x)) / x = x sin(1/x).
Step 5: Now, we need to find the limit of x sin(1/x) as x approaches 0.
Step 6: Use the fact that sin(1/x) is bounded between -1 and 1. Therefore, x sin(1/x) is bounded between -x and x.
Step 7: As x approaches 0, both -x and x approach 0. By the Squeeze Theorem, the limit of x sin(1/x) as x approaches 0 is also 0.
Step 8: Since the limit exists and equals 0, we conclude that f'(0) exists.
