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January 22, 2017
Posted by **khela** on Thursday, October 28, 2010 at 8:26pm.

lim sin(x-1)/(x^2 + x - 2)

x->1

- maths -
**jai**, Thursday, October 28, 2010 at 9:29pmi 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.