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endeavor
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Determine whether f'(0) exists.
f(x) = { x2sin (1/x) if x does not equal 0, 0 if x = 0}
f'(0) = \lim_{x\rightarrow 0} \frac{x^2 \sin \frac{1}{x}}{x - 0}
f'(0) = \lim_{x\rightarrow 0} \frac{\sin \frac{1}{x}}{\frac{1}{x}}
Let u = 1/x. As x approaches 0, u approaches infinity. So I can't use lim x->0 (sin x)/x = 1. But now I have:
f'(0) = \lim_{u\rightarrow \infty} \frac{sin u}{u}
since sin u oscillates between -1 and 1, and u goes to infinity, f'(0) = 0.
f(x) = { x2sin (1/x) if x does not equal 0, 0 if x = 0}
f'(0) = \lim_{x\rightarrow 0} \frac{x^2 \sin \frac{1}{x}}{x - 0}
f'(0) = \lim_{x\rightarrow 0} \frac{\sin \frac{1}{x}}{\frac{1}{x}}
Let u = 1/x. As x approaches 0, u approaches infinity. So I can't use lim x->0 (sin x)/x = 1. But now I have:
f'(0) = \lim_{u\rightarrow \infty} \frac{sin u}{u}
since sin u oscillates between -1 and 1, and u goes to infinity, f'(0) = 0.
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