i have a test tomorrow, and i want help on this, please

lim x(2^(1/x)-1)

x-> +INFINITY

i know the derivative of x^n = nx^(n-1), but when i do that i get (1/x)2^(1/x - 1)/ (1/x), but what do i do next, use l'hospital's rule again?

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- Oct 15th 2008, 07:27 PMskabaniL'Hospital's Rule
i have a test tomorrow, and i want help on this, please

lim x(2^(1/x)-1)

x-> +INFINITY

i know the derivative of x^n = nx^(n-1), but when i do that i get (1/x)2^(1/x - 1)/ (1/x), but what do i do next, use l'hospital's rule again? - Oct 15th 2008, 08:16 PMterr13
In you example, n is not a variable but a constant, so the same cannot be said. The derivative of 2^(1/x) is 2^(1/x)ln(1/x)(-1/2(x^2)).

The derivative of a^x = (a^x)ln(x) and then using chain rule we get the derivative of x^-1 = -(x^-2)/2. - Oct 15th 2008, 08:24 PMskabani
okay, but when i take the limit of that, over (1/x) i get 0/0 , which is still indeterminite...