#Limits
32 messages · Page 1 of 1 (latest)
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you made a mistake
5^(1/n) - 3^(1/n) = (5^(1/n)-1) - (3^(1/n)-1)
so there should not be a +2
otherwise it seems good
@crude rune
-3^1/n-1 is not equal to -(3^(1/n)-1)
how can I substract 1 from each one and not add 2 as a counter
that's your mistake
you're already substracting
5^(1/n) - 3^(1/n) = 5^(1/n) - 3^(1/n) + 1 - 1 = 5^(1/n) - 1 - 3^(1/n) + 1 = (5^(1/n)-1) - (3^(1/n)-1)
Oh I get it now
one last thing: can you justify the manipulation you're doing on line 3 ?
It made more sense at the beginning of the limit to just add 2 right away
(5^(1/n) - 1) / (1/n) = ln(5) form the formula
this point is ok
I'm more concerned about lim n × un = lim n × vn
unless you know equivalents and you can justify that un ~ vn, it would be better to distribute the n× right away
you're welcome, you can close the thread with +close if you're finished
+close
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