Evaluate #lim_(x->0)(sinx/x)^(sinx/(x-sinx))# ?

1 Answer
Jan 23, 2018

#1/e#

Explanation:

With weird limits like this, a good way to handle them is through series expansion.

For small #absx# we have

#sin x = x -x^3/(3!)+O(x^5)# then

#sinx/x = (x -x^3/(3!)+O(x^5))/x = 1-x^2/(3!) + O(x^4) #

and

#sinx/(x-sinx) = (x - x^3/(3!)+O(x^5))/(x-x+x^3/(3!)-O(x^5))=(1-x^2/(3!) + O(x^4))/(x^2/(3!)-O(x^4))#

and then for small #abs x#

#(sinx/x)^(sinx/(x-sinx)) approx (1-x^2/(3!))^((1-x^2/(3!) +)/(x^2/(3!)))#

now calling #x^2/(3!) = 1/y#

#(sinx/x)^(sinx/(x-sinx)) approx (1-1/y)^((1-1/y)/(1/y)) = (1-1/y)^(y-1)#

now #lim_(x->0)(sinx/x)^(sinx/(x-sinx)) equiv lim_(y->oo)(1-1/y)^(y-1)# and

# lim_(y->oo)(1-1/y)^(y-1) = (lim_(y->oo)(1-1/y)^y)/(lim_(y->oo)(1+1/y)) = e^-1/1 = e^-1#