Posted by khela on Thursday, October 28, 2010 at 8:26pm.
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.