limit as x->1/3 of (2-6x)^2/(3x-1)(9x^2-1)
The Attempt at a Solution
substituting gets 0/0, tried expanding but it doesnt work either, dunno what else is there left to do...
No, this should be 4(3x - 1)(3x -1).simplifying the numerator gives 36x^2-24x+4
factor it to (12x+4)(3x+1)
factor it again to 4(3x+1)(3x+1)
Can you cancel 3 + 1 and 3 - 1?so
can i cancel (3x+1) and (3x-1) ?
well i tried canceling it anyway and then substituted 1/3 again now i get 8/0 so its +infinity i guess.
yea i got 4(9x^2 - 6x + 1) the first time as well but didnt know what to do with it further.. im bad at factoringactually i factored out the numerator a different way
i took out the 4 from the beginning and got 4(9x^2 - 6x + 1) then got (3x-1)(3x-1)
you neglected the fact that it -24x, not +24x when you factored it
for the denominator: (3x-1)(9x^2 - 1)
recognize the difference of squares in the denominator?
it will become(3x-1)(3x-1)(3x+1), so cancel and evaluate
just curious do you know about l'hopital's rule?
if you do, try to evaluate the limit using that rule and you see that the answer is different
just another example of when l'hopital's rule fails if you're interested in knowing that
yea i got it thanks!I hope you realized that when I asked "can you cancel 3 + 1 and 3 - 1" you understood me to mean that you can't do that. And you can't cancel 3x + 1 and 3x - 1.
So what did you get as your final expression before you took the limit?