Have you tried L'hospital's rule?
L'Hôpital's rule - Wikipedia, the free encyclopedia
Find the solution for: [(1+cx)^1/3 - 1]/x as x approaches 0.
Hello, I've been trying to solve this problem for a little over a month without much progress..
Looking at it, the logical answer seems to be to get the (1+cx) out of it's third root form to cancel the constants and remove x, but I can't seem to do that w/o adding other similar terms. I've tried adding elements to the top to get a factorial out but that hasn't worked, I've tried multiplying by the positive counterpart ((1+cx)^1/3 + 1) to up the polynomial without luck, and lately I've been playing with natural logs to bring the 1/3 down; all to no avail.
I know the limit exists for sure, any help would be appreciated.
Thanks,
Rhek
Have you tried L'hospital's rule?
L'Hôpital's rule - Wikipedia, the free encyclopedia
I've never like using "L'Hopital" if a limit could be done without Calculus.
Recall that
Taking y= 1,
Replacing x by , and and
That is, can be written as
And, the limit of that, as x goes to 0, is easy.
Thanks for both the responses guys. I should have stipulated that the problem is meant to be done w/o the use of derivatives, but now I/ve got both solutions. Thanks so much for the help, been scratching my head over that one for a long time.
-Rhek