lim (x--pi/2) (pi-2x)^cosx
To find the limit of the function (pi-2x)^cosx as x approaches pi/2, we can start by noticing that the function contains a composition of two functions: (pi-2x) and cosx.
First, let's analyze the inner function, (pi-2x). As x approaches pi/2, the expression pi-2x approaches pi-2(pi/2) = pi-2(pi/2) = pi-π = 0.
Now, let's focus on the outer function, cosx. As x approaches pi/2, cosx approaches cos(pi/2) = 0.
Therefore, the overall function (pi-2x)^cosx approaches 0^0, which is an indeterminate form. In this case, we cannot determine the exact value of the limit without further analysis or applying additional mathematical techniques.
To further analyze this limit, we can rewrite the original function using the exponential form:
(pi-2x)^cosx = e^(cosx * ln(pi-2x))
Now, using the well-known limit property, we have:
lim (x->a) [f(x)]^g(x) = e^(lim (x->a) [g(x)]*[ln(f(x))])
Applying this property to our function:
lim (x->pi/2) (pi-2x)^cosx = e^(lim (x->pi/2) [cosx * ln(pi-2x)])
Since we have already determined that cosx approaches 0 as x approaches pi/2, the limit simplifies to:
lim (x->pi/2) [cosx * ln(pi-2x)] = 0 * ln(0) = 0
Therefore, the limit of (pi-2x)^cosx as x approaches pi/2 is 0.
if x --> π/2 , then cosπ/2 ---> 0
anything^0 = 1
so lim (π - 2x)^cosx , as x--->π/2
= 1
However, pi-2x->0 as well, and 0^0 is undefined
Haven't figured out yet the proof. L'Hospital's Rule seems not to help much, taking logs.