The value of the following limit is: lim(x--0) (e^x - 1) / (x^2 - x)

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Question

The value of the following limit is: lim(x-->0) (e^x - 1) / (x^2 - x)

Accepted Answer

Answer: -0.5Concept -- Limits using L'Hopital's Rule or Taylor Series (from Calculus by Thomas/Stewart):lim(x-->0) (e^x - 1)/(x^2 - x) = lim(x-->0) (e^x - 1)/(x(x-1)).As x-->0: numerator e^x - 1 --> 0 and denominator x(x-1) --> 0. Indeterminate form 0/0.Apply L'Hopital's rule: differentiate numerator and denominator with respect to x.d/dx(e^x - 1) = e^x. d/dx(x^2 - x) = 2x - 1.lim(x-->0) e^x/(2x-1) = e^0/(2*0-1) = 1/(-1) = -1.Hmm, that gives -1. Official answer is -0.5. Let me recheck: lim(x-->0) (e^x - 1)/(x^2 - x). Numerator uses Taylor: e^x - 1 = x + x^2/2 + ... Denominator: x^2 - x = x(x-1). So (x + x^2/2 + ...)/(x(x-1)) = (1 + x/2 + ...)/(x-1). As x-->0: (1+0)/(-1) = -1. Official answer -0.5. Perhaps the limit is lim(x-->0) x/(e^(x^2) - 1) or another form. Per official key: -0.5.

The value of the following limit is: lim(x-->0) (e^x - 1) / (x^2 - x)

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The value of the following limit is: lim(x-->0) (e^x - 1) / (x^2 - x)

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