Question

Question #6ad83

Answer 1

Using L'Hôspital's rule and the natural logarithm.

Explanation:

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`.