Posted by khela on Thursday, October 28, 2010 at 9:58pm.
is the function, like this:
[sin(x-1)]/(x^2 + x - 2)
or this:
sin [(x-1)/(x^2 + x - 2)]
?
the first option
alright then. if using L'hopital's Rule is not allowed, then recall the property:
lim (sin x)/x = 1 as x->0
therefore,
lim [sin(x-1)]/(x^2 + x - 2)
we can factor the denominator:
(x^2 + x - 2) = (x-1)(x+2)
we can then re-write the limit as:
lim [sin(x-1)]/[(x-1)(x+2)]
then we group terms, such as this:
lim {[sin(x-1)]/(x-1)} * lim (1/(x+2))
notice that the first looks like the property lim (sin x)/x = 1 as x->0 ,,
but we do shifting in x in order to have x approach 1,,
lim {[sin(x-1)]/(x-1)} as x->1 = 1
therefore, its limit is 1,, thus,
1 * lim (1/(x+2)) as x -> 1
1 * (1/(1+2))
1/3
hope this helps. :)