Calculus

Use pinching theorem to evaluate lim x-->1 ((x-1)sin(1/x-1))

I'm confused in the pinch theorem analytically

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1. note that if u = 1/(x-1) then what you have is

sin(u)/u

You have probably seen that this limit is 1, so follow the same argument.

Or, try google. A good discussion is at

math.ucsb.edu/~jcs/SqueezeTheorem.pdf

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2. Would it be 0?

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3. you are correct. 1/(x-1) -> ∞

I tried to pound a round peg into a square hole.

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