Question

I don't see the reasoning in going straight from this limit to the answer is one.. someone help.

(((1+n^(1/7)+n^(1/6))/(2-n+n^(4/3))) *

(n^(7/6)) / 1

Answer 1

\[((1+ \sqrt[7]{n}+\sqrt[6]{n})\div(2-n+ \sqrt[3]{n ^{4}}))(\sqrt[6]{n ^{7}})\]

Written out a little nicer

Answer 2

Do u need to calculate limit?

Answer 3

Well I'm doing a limit comparison test, and they went straight from this mess to the limit equals one.. I just need to know this limit doesn't equal 0 really..

Answer 4

And it's as it goes to infinity

Answer 5

http://www.wolframalpha.com/input/?i=lim+n+-%3E+infinity+%28%28%281%2Bn^%281%2F7%29%2Bn^%281%2F6%29%29%2F%282-n%2Bn^%284%2F3%29%29%29+*+%28n^%287%2F6%29%29

Answer 6

check it out

Answer 7

I already wolframmed it.. I clicked show steps and there were none, lol. but i'll check it out

Answer 8

ok..

Answer 9

yeah.. still no steps

Answer 10

Lim comparison test, they took the function they were evaluating, they took another function that they know how it behaves, function a/function b take lim. Limit is 1is helpful info: lim>0 says function a and function b either both converge or both diverge.

Answer 11

Yeah, I need help understanding why exactly that limit it one. It looks like there is too much going on to start L'hopitaling it.. And the top and both both have different powers when multiplied through by n^(7/6)

Answer 12

For better readability:

\[\frac{1 + n^{1/7} + n^{1/6}}{2-n +n^{4/3}} * \frac{n^{7/5}}{1}\]

Answer 13

Is that right?

Answer 14

Simplify, take lim of n^(highest exponent)/n^(highest exponent)

Answer 15

yes except it's n^(7/6) / 1 at the end

Answer 16

Ok, so distribute that n through we get

\[\frac{n^{7/6} + n^{1/7 + 5/7} + n^{1/6 + 5/7}}{2-n + n^{4/3}}\]

Answer 17

Err that's wrong.

Answer 18

Got my fractions upside-down
\[\frac{n^{7/6} + n^{1/7 + 7/6} + n^{1/6 + 7/6}}{2-n+n^{4/3}}\]

Answer 19

lol.. thats just perfect isn't it

Answer 20

oops...

Answer 21

I was trying to multiply the exponents.. I see why it's one now

Answer 22

So on top we have the largest power of n being \( n^{8/6}\)

Answer 23

And on bottom, same thing.

Answer 24

So yeah, 1.

Answer 25

:P

Answer 26

When you multiply powers of the same base you add their exponents.

Answer 27

yeah....