Prove that the function f(x) = x sin(1/x) for x ≠ 0 and f(0) = 0 is continuous everywhere but not differentiable at x=0.
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Step-by-Step Solution
Step 1: Check continuity for x=0
For x=0, the function f(x)=xsin(1/x) is a product of two continuous functions: g(x)=x and h(x)=sin(1/x). The function g(x)=x is continuous everywhere. The function h(x)=sin(1/x) is a composition of sin(u) and u=1/x. Since 1/x is continuous for x=0 and sin(u) is continuous everywhere, their composition is continuous for x=0. Therefore, their product f(x) is continuous for all x=0.
Step 2: Check continuity at x=0
To check continuity at x=0, we need to evaluate the limit of f(x) as x approaches 0. We know that the sine function is bounded between −1 and 1, i.e., −1≤sin(1/x)≤1. Multiplying this inequality by ∣x∣ (which is positive), we get −∣x∣≤xsin(1/x)≤∣x∣. As x approaches 0, ∣x∣ approaches 0. By the Squeeze Theorem, limx→0xsin(1/x)=0. Since f(0) is given as 0, we have limx→0f(x)=f(0), which means the function is continuous at x=0.
Step 3: Check differentiability at x=0
To check differentiability at x=0, we use the definition of the derivative: f′(0)=limh→0hf(0+h)−f(0). Substituting the function definition, we get limh→0hhsin(1/h)−0. This simplifies to limh→0sin(1/h).
Step 4: Evaluate the limit for differentiability
As h approaches 0, 1/h approaches ±∞. The function sin(u) oscillates between −1 and 1 as u approaches ±∞. Therefore, limh→0sin(1/h) does not approach a single value and thus does not exist. Since the limit does not exist, the function is not differentiable at x=0.