anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[((1+ \sqrt[7]{n}+\sqrt[6]{n})\div(2-n+ \sqrt[3]{n ^{4}}))(\sqrt[6]{n ^{7}})\] Written out a little nicer
anonymous
  • anonymous
Do u need to calculate limit?
anonymous
  • 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..

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More answers

anonymous
  • anonymous
And it's as it goes to infinity
anonymous
  • anonymous
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
anonymous
  • anonymous
check it out
anonymous
  • anonymous
I already wolframmed it.. I clicked show steps and there were none, lol. but i'll check it out
anonymous
  • anonymous
ok..
anonymous
  • anonymous
yeah.. still no steps
anonymous
  • 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.
anonymous
  • 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)
anonymous
  • anonymous
For better readability: \[\frac{1 + n^{1/7} + n^{1/6}}{2-n +n^{4/3}} * \frac{n^{7/5}}{1}\]
anonymous
  • anonymous
Is that right?
anonymous
  • anonymous
Simplify, take lim of n^(highest exponent)/n^(highest exponent)
anonymous
  • anonymous
yes except it's n^(7/6) / 1 at the end
anonymous
  • anonymous
Ok, so distribute that n through we get \[\frac{n^{7/6} + n^{1/7 + 5/7} + n^{1/6 + 5/7}}{2-n + n^{4/3}}\]
anonymous
  • anonymous
Err that's wrong.
anonymous
  • anonymous
Got my fractions upside-down \[\frac{n^{7/6} + n^{1/7 + 7/6} + n^{1/6 + 7/6}}{2-n+n^{4/3}}\]
anonymous
  • anonymous
lol.. thats just perfect isn't it
anonymous
  • anonymous
oops...
anonymous
  • anonymous
I was trying to multiply the exponents.. I see why it's one now
anonymous
  • anonymous
So on top we have the largest power of n being \( n^{8/6}\)
anonymous
  • anonymous
And on bottom, same thing.
anonymous
  • anonymous
So yeah, 1.
anonymous
  • anonymous
:P
anonymous
  • anonymous
When you multiply powers of the same base you add their exponents.
anonymous
  • anonymous
yeah....

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