|jarmen||x^(1/x) seems to go to 1|
|jarmen||damnit, i cant see how that should go to 3|
|action||jarmen opens up mathematica|
|jarmen||opens up mathematica|
|action||davo checks a solutions manual for other ideas|
|davo||checks a solutions manual for other ideas|
|jarmen||mathematica says the limit is 1|
are you sure the limit is 3?
Limit[Log[2^x + 3^x]^(1/x), x \[Rule] Infinity]
|action||davo checks the answer book again|
|davo||checks the answer book again|
taking for x -> infinity...
I'd get (1/x)*ln(2^x + 3^x)...
|davo||yeah, it says 'lim(n->+inf) [(2^n + 3^n)^(1/n)] = 3|
wheres the log in that expression davo?
|excelblue||by l'hospital's rule, I get (2^x/ln(2), 3^x/ln(3))/1|
for n -> inf, that is inf
|jarmen||excelblue, how did you just bring the (1/x) down|
|excelblue||substituting back -> e^inf = inf|
|excelblue||jarmen: look at the original problem|
lim x->inf (2^x + 3^x)^(1/x)
that's the original
|jarmen||wtf, then what is this problem: ln(2^x + 3^x)^(1/x)|
|excelblue||I put the parenthesis wrong|
when I told him to take ln
and his original problem shifted
|excelblue||% Limit[(2^x + 3^x)^(1/x), x, Infinity]|
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]|