Determine whether f'(0) exists.(adsbygoogle = window.adsbygoogle || []).push({});

f(x) = { x^{2}sin (1/x) if x does not equal 0, 0 if x = 0}

[tex]f'(0) = \lim_{x\rightarrow 0} \frac{x^2 \sin \frac{1}{x}}{x - 0}[/tex]

[tex]f'(0) = \lim_{x\rightarrow 0} \frac{\sin \frac{1}{x}}{\frac{1}{x}}[/tex]

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:

[tex]f'(0) = \lim_{u\rightarrow \infty} \frac{sin u}{u}[/tex]

since sin u oscillates between -1 and 1, and u goes to infinity, f'(0) = 0.

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# Homework Help: Is this right? (limits w/ trig)

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