How do I solve this limit ?

#lim_(x->0) ((1+ax)^(1/x)-(1+x)^(a/x))/x#

1 Answer
Apr 22, 2018

#e^a*(a/2)*(1 - a)#

Explanation:

#"You could use Taylor series and drop higher order terms in the"# #"limit for "x->0"."#

#x^y = exp(y*ln(x))#
#=> (1+x)^y = exp(y*ln(1+x))#
#"and "ln(1+x) = x - x^2/2 + x^3/3 - ...#
#"and "exp(x) = 1 + x + x^2/2 + x^3/6 + x^4/24 + ...#
#"So"#
#exp(y*ln(1+x)) = exp(y*(x - x^2/2 + ...))#

#=> (1+x)^(a/x) = exp((a/x)*ln(1+x))#

#= exp((a/x)*(x - x^2/2 + x^3/3 - ...))#

#= exp(a - a*x/2 + a*x^2/3 - ...)#

#=> (1+ax)^(1/x) = exp((1/x)*ln(1+ax))#

#= exp((1/x)*(ax - (ax)^2/2 + (ax)^3/3 - ...))#

#= exp(a - a^2*x/2 + a^3*x^2/3 - ...)#

#=>(1+ax)^(1/x) - (1+x)^(a/x)#

#~~ exp(a - a^2*x/2 + ...) - exp(a - a*x/2 + ...)#

#~~ exp(a)/exp(a^2*x/2) - exp(a)/exp(a*x/2)#

#= exp(a) (exp(-a^2*x/2) - exp(-a*x/2))#

#~~ exp(a) (1 - a^2*x/2 - 1 + a*x/2)#

#= exp(a) ((x/2)(a - a^2))#

#=>((1+ax)^(1/x) - (1+x)^(a/x))/x#

#~~ exp(a) ((1/2) (a - a^2))#

#= e^a*(a/2)*(1 - a)#