Posted by khela on Thursday, October 28, 2010 at 9:58pm.
find the limit of lim sin(x1)/(x^2 + x  2) x>1
without using l'hopitals rule

maths  jai, Thursday, October 28, 2010 at 10:01pm
is the function, like this:
[sin(x1)]/(x^2 + x  2)
or this:
sin [(x1)/(x^2 + x  2)]
? 
maths  khela, Thursday, October 28, 2010 at 10:10pm
the first option

maths  jai, Thursday, October 28, 2010 at 10:40pm
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(x1)]/(x^2 + x  2)
we can factor the denominator:
(x^2 + x  2) = (x1)(x+2)
we can then rewrite the limit as:
lim [sin(x1)]/[(x1)(x+2)]
then we group terms, such as this:
lim {[sin(x1)]/(x1)} * 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(x1)]/(x1)} 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. :)