Evaluate the limit #lim_(x→0)(sin x/x)^(1/x^2)#?

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Answer 1 · 2016-10-16T15:52:08

#e^(-1/6)#

#sinx = x-x^3/(3!)+x^5/(5!)+x^7/(7!)+ cdots#

this is an alternate series and if #abs x < 1#

#(x-x^3/(3!))/x < sin x/x < (x-x^3/(3!)+x^5/(5!))/x#

#lim_(x->0)((x-x^3/(3!))/x)^(1/x^2)=lim_(x->0)(1-x^2/6)^(1/x^2)#

but making #y = -x^2/6#

#lim_(x->0)(1-x^2/6)^(1/x^2) =lim_(y->0)((1+y)^(1/y))^(-1/6)=e^(-1/6) #

analogously

#lim_(x->0)( (x-x^3/(3!)+x^5/(5!))/x)^(1/x^2) = lim_(x->0) (1-x^2/(3!)+x^4/(5!))^(1/x^2) = e^(-1/6)#