Final answer:
The limit of (sin(x))^x as x approaches 0+ is found using L'Hôpital's Rule after transforming the expression with a natural logarithm. By applying L'Hôpital's Rule, we find that the limit approaches 1.
Step-by-step explanation:
The student is asking about the limit of (sin(x))^x as x approaches 0+. To solve this limit, we must first recognize that as x approaches 0, sin(x) approaches 0, making the direct computation of the limit difficult since the base is going to zero while the exponent is also going to zero. We can address this indeterminacy by taking the natural logarithm of the function, applying L'Hôpital's Rule, and then exponentiating the result.
Let L be the limit we want to find:
L = lim (x -> 0+) (sin(x))^x
Take the natural log of both sides to transform the expression into a product:
ln(L) = lim (x -> 0+) x*ln(sin(x))
This expression is now in a 0*infinity form, another indeterminate form. Now, we can use L'Hôpital's Rule by rewriting x as 1/x^(-1) and considering this a product of two functions.
After applying L'Hôpital's Rule and simplifying, we find that ln(L) approaches 0. Thus, L itself must approach 1. We can conclude that the limit of (sin(x))^x as x approaches 0+ is 1.