Question

lim as (x->1) of (x^(1/3)-1)/((sqrt of x) -1)

Answer 1

\[\large \lim_{x \to 1} \frac{x^{\frac{1}{3}}-1}{\sqrt{x}-1} \]
this one?

Answer 2

multiply by the conjugate

Answer 3

actually I think you need to be a bit more tricky and use difference of squares on the top, and difference of cubes on the bottom

Answer 4

@Spacelimbus Yes
@wio That's the problem there, I get (x^(5/6)-1)/((x-1)

Answer 5

I believe best is to combine what @wio did with the rule of Bernoulli de L'Hopital, one differentiation should be enough. If you're allowed to use that yet.

Answer 6

@Spacelimbus my professor didn't even teach us that, and refers us to it. :(

Answer 7

I promise you, what I said works

Answer 8

difference of squares on top, difference of cubes on the bottom. cancellation happens...

Answer 9

I don't quite understand @LostAnkh, he did not teach your Bernoulli de l'Hopital and wants you to use it? Or you haven't been taught about limits?

Answer 10

@Spacelimbus Sorry, I meant, he didn't teach us L'Hopital's rule, but refers to it.

Answer 11

\[\large \lim_{x \to 1} \frac{x^{1/3}-1}{x^{1/2}-1}=\lim_{x \to 1} \frac{(x^{1/6}-1)(x^{1/6}+1)}{(x^{1/6}-1)(x^{1/3}+x^{1/6}+1)}\]

Answer 12

tadah!

Answer 13

What @TuringTest recommends will work without L'Hopital, but might be not the first intuitive thing you spot on first glance, so I understand your Professor in this manner, however you shouldn't be linked to stuff you haven't been taught yet in class.

If you want to learn about the simple computation of L'Hopital I recommend you to search a bit on the internet, not because I am too lazy to help you with it, but because the material you will find there is much better than my explanation could be.

Answer 14

@TuringTest That works and gets the answer correct, but I wouldn't have thought to differentiate by cube for the bottom.
@Spacelimbus understandable.

Answer 15

once I realized the common factor in the exponent was 1/6, then realized the top could be treated as a difference of squares, naturally the only thing to do was difference of cubes on the bottom. The "1"'s guided me, as I noticed they are perfect squares, cubes, etc... in case you want my thought process

Answer 16

I'm not saying it should jump out at you, but keep that goal of cancellation in mind and don't forget your algebra!

Answer 17

@TuringTest Thank you.

Answer 18

Welcome!