At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
dude none of them are going to help you
:( maybe you're right. 3
Don't be so pessimistic. Did you try actually dividing those polynomials as far as you can?
you have to divide it so that when you plug x=3 , the result is not 0/0
i still get 0/0 guys
by using L hopital's rule i get 14/13. but the question must be solved w/o using it. :(
usually, there will be a common factor between the denominator and the numerator, but i can't factorize those polynomial
this problem sure is hard. xD
If a polynomial has a root "a", there is a factor (x-a) in it!
i don't get your point bro.
i got the common factor ! it's gonna be -> (x-3)(x^2 + 3x -4) / (x-3)(x^2+3x-5)
you can eliminate x-3, then plug x=3 you'll get -> (9+9-4)/(9+9-5) = 14/13
whoa! thanks din!!!
you're welcome btw, i got the common factor by using Horner's method
I'm not familiar with it. can you please give some background :D
i'll just google it. :)) thanks din for the info. ;)
lol! it's just synthetic division :))
haha in my school we usually call it horner's method
The denominator and the numerator are both 0 when x = 3, what does that tell you?
it means that the function is in indeterminate form :)
obtaining 0/0 in a function doesn't that it has no limit. :)
what the answer in this question? please. I help me out 😉