#math - Sun 18 Mar 2007 between 00:00 and 00:06

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jarmenx^(1/x) seems to go to 1
seems reasonable
davoyep
jarmendamnit, i cant see how that should go to 3
actionjarmen opens up mathematica
jarmenopens up mathematica
actiondavo checks a solutions manual for other ideas
davochecks a solutions manual for other ideas
jarmenmathematica says the limit is 1
are you sure the limit is 3?
davowow really?
jarmenIn[1]:=
Limit[Log[2^x + 3^x]^(1/x), x \[Rule] Infinity]
Out[1]=
1
actiondavo checks the answer book again
davochecks the answer book again
excelbluelet's see...
taking for x -> infinity...
I'd get (1/x)*ln(2^x + 3^x)...
davoyeah, it says 'lim(n->+inf) [(2^n + 3^n)^(1/n)] = 3
jarmenuh..
wheres the log in that expression davo?
excelblueby l'hospital's rule, I get (2^x/ln(2), 3^x/ln(3))/1
for n -> inf, that is inf
jarmenexcelblue, how did you just bring the (1/x) down
excelbluesubstituting back -> e^inf = inf
fezhi
excelbluejarmen: look at the original problem
lim x->inf (2^x + 3^x)^(1/x)
that's the original
jarmenwtf, then what is this problem: ln(2^x + 3^x)^(1/x)
excelblueI put the parenthesis wrong
when I told him to take ln
and his original problem shifted
jarmenoh
excelblue% Limit[(2^x + 3^x)^(1/x), x, Infinity]
mbotexcelblue:
x x 1/x
Limit::nonopt: Options expected (instead of Infinity) beyond position 2 in Limit[(2 + 3 ) , x, Infinity]. An option must be a rule or a list of rules.
Limit[(2^x + 3^x)^x^(-1), x, Infinity]
excelblue% Limit[(2^x + 3^x)^(1/x), x->Infinity]
mbotexcelblue: 3

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