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

### NY Lost Funds

 jarmen x^(1/x) seems to go to 1seems reasonable davo yep 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 1are you sure the limit is 3? davo wow really? jarmen In[1]:=Limit[Log[2^x + 3^x]^(1/x), x \[Rule] Infinity]Out[1]=1 action davo checks the answer book again davo checks the answer book again excelblue let's see...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 jarmen uh..wheres the log in that expression davo? excelblue by l'hospital's rule, I get (2^x/ln(2), 3^x/ln(3))/1for n -> inf, that is inf jarmen excelblue, how did you just bring the (1/x) down excelblue substituting back -> e^inf = inf fez hi excelblue jarmen: look at the original problemlim 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 wrongwhen I told him to take lnand his original problem shifted jarmen oh excelblue % Limit[(2^x + 3^x)^(1/x), x, Infinity] mbot excelblue:x x 1/xLimit::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] mbot excelblue: 3