Question

another limits question, please help!!
will be typed in equation editor in a few mins.

Answer 1

ok

Answer 2

\[\lim_{x \rightarrow 64} [x-64]\div \sqrt[3]{x}-4\]

Answer 3

is that 3rd root of x?

Answer 4

yes

Answer 5

see here we rationalize the denom to get rid of that 3rd root

Answer 6

okay

Answer 7

or actually, we can use hopitals rule...

Answer 8

i don't know what that is :/

Answer 9

when u have a limit of the form 0/0 by plugging in the limiting value, we can take the derivative of the top and derivative of the bottom and re-write the fraction and do the limit again

Answer 10

in this case:
= lim x--> 64 (1 / (1/3)x^(-2/3)) and by plugging in 64 (bcuz we can now) we get:

= 48

Answer 11

oh, i havent learned it that way, so i don't know if it would be accepted on my test..
do you mind showing me how to do it normally?

Answer 12

ok well the quick way is to simplify x-64 as a difference of cubes:

\[\left( x-64 \right)=(\sqrt[3]{x}-4)(x^{2/3}+4\sqrt[3]{x}+16)\]

and then plug in 64

Answer 13

since the denominator will cancel out leaving the trinomial...