jarmen | x^(1/x) seems to go to 1 seems 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 1 are 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))/1 for 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 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 |

jarmen | oh |

excelblue | % Limit[(2^x + 3^x)^(1/x), x, Infinity] |

mbot | excelblue: 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] |

mbot | excelblue: 3 |