Evaluate _x 0^+ x^ x

Here are the evaluations for the given limits: 1) Evaluate _x 0^+ x^ x Step 1: Identify the indeterminate form. As x 0^+, x 0 and x 0. This is an indeterminate form of type 0^0. Step 2: Use logarithms to convert to a product form. Let y = x^ x. Then y = (x^ x) = x x. Now, we evaluate _x 0^+ y = _x 0^+ x x. This is an indeterminate form of type 0 · (-). Step 3: Rewrite the expression to apply L'Hôpital's Rule. Rewrite x x as ( x)/(1/ x) = ( x)/( x). Now, as x 0^+, x - and x . This is an indeterminate form of type (-)/(). Step 4: Apply L'Hôpital's Rule. _x 0^+ (d)/(dx)( x)(d)/(dx)( x) = _x 0^+ (1/x)/(- x x) = _x 0^+ (1)/(x) · (- x)/( x) · x = _x 0^+ (-^2 x)/(x x) We can rewrite this as: = _x 0^+ (-( x)/(x) · ( x)/( x)) = _x 0^+ (-( x)/(x) · x) Step 5: Evaluate the limit. We know that _x 0^+ ( x)/(x) = 1 and _x 0^+ x = 0. So, _x 0^+ y = -(1 · 0) = 0. Step 6: Find the original limit. Since _x 0^+ y = 0, then _x 0^+ y = e^0 = 1. The limit is 1. 2) Evaluate _x 0^+ (1 + 3x)^1/(2x) Step 1: Identify the indeterminate form. As x 0^+, 1 + 3x 1 and (1)/(2x) . This is an indeterminate form of type 1^. Step 2: Use the standard limit definition of e. Recall the definition: _u 0 (1 + u)^1/u = e. Let u = 3x. As x 0^+, u 0^+. Substitute u into the expression: _u 0^+ (1 + u)^(1)/(2(u/3)) _u 0^+ (1 + u)^(3)/(2u) Step 3: Manipulate the exponent to match the definition of e. _u 0^+ [(1 + u)^(1)/(u)]^(3)/(2) Step 4: Evaluate the limit. Using the property of limits for powers: = [_u 0^+ (1 + u)^(1)/(u)]^(3)/(2) = e^(3)/(2) The limit is e^3/2. 3) Evaluate _x (x^2 + x)^1/ x Step 1: Identify the indeterminate form. As x , x^2 + x and x , so (1)/( x) 0. This is an indeterminate form of type ^0. Step 2: Use logarithms to convert to a product form. Let y = (x^2 + x)^1/ x. Then y = ((x^2 + x)^1/ x) = (1)/( x) (x^2 + x) = ((x^2 + x))/( x). Now, we evaluate _x y = _x ((x^2 + x))/( x). As x , (x^2 + x) and x . This is an indeterminate form of type ()/(). Step 3: Apply L'Hôpital's Rule. _x (d)/(dx)((x^2 + x))(d)/(dx)( x) = _x (2x+1)/(x^2+x)(1)/(x) Step 4: Simplify the expression. = _x (2x+1)/(x^2+x) · x = _x (x(2x+1))/(x^2+x) = _x (2x^2+x)/(x^2+x) Step 5: Evaluate the limit by dividing by the highest power of x in the denominator. Divide the numerator and denominator by x^2: = _x (2x^2)/(x^2)+(x)/(x^2)(x^2)/(x^2)+(x)/(x^2) = _x (2+1)/(x)1+(1)/(x) As x , (1)/(x) 0. = (2+0)/(1+0) = 2 So, _x y = 2. Step 6: Find the original limit. Since _x y = 2, then _x y = e^2. The limit is e^2. That's 2 down. 3 left today — send the next one.