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Calculus with Analytical geometry

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I'm stumped on this one.

lim |x-1|sin(1/x-1)
x->1

  • Calculus with Analytical geometry -

    as written, there's no problem. Just plug in 1 to get

    |1-1| (sin 1/1-1) = 0.

    However, I think you mean

    |x-1|sin(1/(x-1))

    again, no problem. Since sin is always less than 1, we have 0 * z, where z<1, so zero again.

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