How do you find the limit of #[1+(a/x)]^(bx)# as x approaches infinity using l'hospital's rule?

1 Answer
Eddie
Sep 1, 2016

#= e^(ab)#

Explanation:

#lim_(x to oo) [1+(a/x)]^(bx)#

first you need to force it into an indeterminate form, and there's little to lose in trying to but the stuff in parentheses apart so we do this next step

#lim_(x to oo) exp ( ln [1+(a/x)]^(bx) )#

we can lift the #e# out as it is a continuous function
#exp ( lim_(x to oo) bx ln [1+(a/x)] )#

we also notice that #lim_(x to oo) ln [1+(a/x)] = ln 1 = 0#

so the indeterminate form is

#exp ( lim_(x to oo) (ln [1+(a/x)] )/(1/(bx)))# and we can apply L'Hopital

#= exp ( lim_(x to oo) (-a/x^2 1/[1+(a/x)] )/(-1/(bx^2)))#

#= exp ( lim_(x to oo) ( a b)/(1+(a/x)))#

#= e^(ab)#

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