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)