Evaluate the limit of the function (sin^x (x)-1)/(x-pi/2), if x goes to pi/2?

giorgiana1976 | Student

Let f(x) = (sin x)^x

If x = pi/2 => f(pi/2) = (sin pi/2)^pi/2  = 1

We'll re-write the function whose limit has to be found out:

lim (f(x) - 1)/(x - pi/2)

By definition, the derivative of a function f(x), at the point x = pi/2 is: lim (f(x) - f(pi/2))/(x - pi/2) = f'(pi/2).

We must find out the expression of the 1st derivative of f(x).

We'll take natural logarithms both sides:

ln f(x) = ln [(sin x)^x]

ln f(x) = x*ln (sin x)

We'll differentiate with respect to x both sides:

f'(x)/f(x) = ln (sin x) + x*cos x/sin x

f'(x)/f(x) = ln (sin x) + x*cot x

f'(x) = f(x)*[ln (sin x) + x*cot x]

f'(x) = [(sin x)^x]*[ln (sin x) + x*cot x]

Now, we'll replace x by pi/2:

f'(pi/2) = 1*(ln 1 + pi/2*0)

f'(pi/2) = 1*0

f'(pi/2) = 0

Therefore, the limit of the given function, when x approaches to pi/2, is lim [(sin x)^x - 1]/(x - pi/2) = 0.

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