Show that g′ is discontinuous at x = 0 and that g′ takes on values close to ±1 arbitrarily near 0. When g(x) = $$x^2 sin(1/x)$$
Show that g′ is discontinuous at x = 0 and that g′ takes on values close to ±1 arbitrarily near 0.
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real-analysis
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1What's the point of us doing your homework? – 2012-11-08