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maths

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find the limit of
lim sin(x-1)/(x^2 + x - 2)
x->1

  • maths - ,

    i interpreted this in two ways:
    first,
    lim [sin(x-1)]/(x^2 + x - 2) as x->1
    *since, there is no common factor between numerator and denominator, it's in simplest form,, also you substitute 1 to both [sin(x-1)] and (x^2 + x - 2):
    sin(x-1) = sin (1-1) = 0
    (x^2 + x - 2) = i^2 + 1 -2 = 0
    lim 0/0 becomes indeterminate
    *recall that when you have 0/0, you use L'hopital's Rule to get limit:

    L'hopital's Rule: you get the derivative of both numerator and denominator,, then evaluate it at x,, thus:
    derivative of sin(x-1) = cos(x-1)
    derivative of (x^2 + x - 2) = (2x + 1)

    therefore,
    [cos(x-1)]/(2x + 1) evaluated at x=1
    [cos(1-1)]/(2(1) + 1)
    (cos(0))/(2+1)
    you then solve it.

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