## anonymous 5 years ago 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

1. anonymous

$((1+ \sqrt[7]{n}+\sqrt[6]{n})\div(2-n+ \sqrt[3]{n ^{4}}))(\sqrt[6]{n ^{7}})$ Written out a little nicer

2. anonymous

Do u need to calculate limit?

3. anonymous

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..

4. anonymous

And it's as it goes to infinity

5. anonymous
6. anonymous

check it out

7. anonymous

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

8. anonymous

ok..

9. anonymous

yeah.. still no steps

10. anonymous

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.

11. anonymous

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)

12. anonymous

For better readability: $\frac{1 + n^{1/7} + n^{1/6}}{2-n +n^{4/3}} * \frac{n^{7/5}}{1}$

13. anonymous

Is that right?

14. anonymous

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

15. anonymous

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

16. anonymous

Ok, so distribute that n through we get $\frac{n^{7/6} + n^{1/7 + 5/7} + n^{1/6 + 5/7}}{2-n + n^{4/3}}$

17. anonymous

Err that's wrong.

18. anonymous

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

19. anonymous

lol.. thats just perfect isn't it

20. anonymous

oops...

21. anonymous

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

22. anonymous

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

23. anonymous

And on bottom, same thing.

24. anonymous

So yeah, 1.

25. anonymous

:P

26. anonymous

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

27. anonymous

yeah....