Prove: lim as x goes to 0+ of x^(sinx)=1

To prove that the limit of x^(sinx) as x approaches 0 from the positive side is equal to 1, we need to use the definition of a limit and apply properties of exponential and trigonometric functions.

To start, let's write out the limit expression:

lim (x→0+) x^(sinx)

Now, since sin(x) is bounded between -1 and 1, let's consider three cases:

Case 1: sin(x) > 0
In this case, as x approaches 0 from the positive side, sin(x) also approaches 0. And as sin(x) approaches 0, we know that x^(sinx) approaches 1 because any positive number raised to the power of 0 is equal to 1.

Case 2: sin(x) < 0
Similar to Case 1, as x approaches 0 from the positive side, sin(x) approaches 0. However, in this case, x^(sinx) will still approach 1 because any positive number raised to the power of 0 is equal to 1, regardless of the sign of the base.

Case 3: sin(x) = 0
If sin(x) is exactly 0, then x^(sinx) will still be equal to 1 because any number (including 0) raised to the power of 0 is equal to 1.

Now, no matter which case we consider, we see that as x approaches 0 from the positive side, x^(sinx) approaches 1.

Therefore, we have proven that lim as x approaches 0+ of x^(sinx) = 1.