How do you find the limit of #x^(sin(x))# as x approaches 0?

Question

70889 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-27T19:07:23

#1#

let #L = lim_(x to 0) x^(sin x)#

#implies ln L = ln lim_(x to 0) x^(sin x) #

#= lim_(x to 0) ln x^(sin x)#

#= lim_(x to 0) sinx ln x#

#= lim_(x to 0) (ln x)/(1/(sinx) )#

#= lim_(x to 0) (ln x)/(csc x )#

this is in indeterminate #oo/oo# form so we can use L'Hôpital's Rule

#= lim_(x to 0) (1/x)/(- csc x cot x)#

#=- lim_(x to 0) (sin x tan x)/(x)#

Next bit is unnecessary, see ratnaker-m's note below...

this is now in indeterminate #0/0# form so we can go again

#ln L =- lim_(x to 0) (cos x tan x + sin x sec^2 x)/(1)#

#= - 0#

So:
#L = e^(- 0) = 1#

Answer 2 · 2016-09-27T19:19:09

#1#

#x^sin x = (1+x-1)^sinx = 1 +sin x/(1!)(x-1)+(sinx(sinx-1))/(2!)(x-1)^2+cdots+#

then

#lim_(x->0)x^sinx = 1#