Question #6ad83

Let's break the limit into the two parts.

a. First part : #sin^(1/x) x#

Let #lim_(x->0) sin^(1/x) x# be equal to #a#.
Take the natural logarithm of both sides.
#lim_(x->0) 1/x ln(sin x) = ln a#.

Now, in order to simplify things, let's use #x# instead of #sin x#, as #lim_(x->0) sin x = x#. That is totally another thing and will complicate this explanation, so just watch this video for that :
https://www.khanacademy.org/math/ap-calculus-ab/ab-derivative-rules/ab-derivtive-rules-opt-vids/v/sinx-over-x-as-x-approaches-0.

So we have
#lim_(x-> 0) ln x/x = ln a#, however, since #lim_(x->0) ln x# approaches negative infinity, then #lim_(x->0) ln x/x# also approaches negative infinity. Then we raise #e# to both sides and get
#e^ln a = e^("-infinity") -> a = 0.# We're already halfway through.

b. Second part : #(1/x)^(sin x)#

We'll do the same as last time;
Let #lim_(x->0) (1/x)^(sin x) = b#.
Take the natural logarithm of both sides:
#lim_(x->0) -sinx*ln x= ln b#
#lim_(x->0) -x*ln x = ln b#
#lim_(x->0) ln x/(1/-x) = ln b#
This is an infinity/infinity situation, so we can use L'Hôspital's rule; take the derivative of both the numerator and denominator and we get :
#lim_(x->0) (1/x)/(1/x^2) = lim_(x->0) x = 0#.
Raise #e# to both sides again;
#e^ln b = e^0 -> b = 1#.

c. Our original expression is equal to #a+b#, which is equal to #0+1=1#.

So, in conclusion (or tl;dr):
#lim_(x->0) (sin x)^(1/x) + (1/x)^(sin x) = 1#.