Evaluate the limit using l'hopital's rule

Question

Iim (tanx)^π/2-x

x→π/2^-

1. Lim x tends to π/2 (negative)
2. (tanx)^π/2-x

Answer 1

Let's evaluate the limit for this exponential expression.

Simplify the problem.

The new variable y = π/2 - x. Then tan x = tan (π/2 - y) = cot y

Iim (tan x)^π/2-x = lim (cot y)^y

x→π/2^- y→0⁺

New functions f(y) = (cot y)^y and g(y) = ln(f(y)) = y ln(cot y) = y ln(cos y) - y ln(sin y)

lim g(y) = 0 ln(cos 0) + lim [(ln sin y)/(-1/y)] = lim [(ln sin y) / (-1/y)]

y→0⁺ y→0⁺ y→0⁺

Because lim (ln sin y) = -∞ and lim (-1/y) = -∞, we can use l'hopital's rule.

y→0⁺ y→0⁺

lim g(y) = lim [(ln sin y)' / (-1/y)']

y→0⁺ y→0⁺

(ln sin y)' = (1/sin y) (sin y)' = cos y / sin y , (-1/y)' = 1/y²

So, lim g(y) = lim ((cos y / sin y) / 1/y²) = lim (y • cos y • y/sin y)

y→0⁺ y→0⁺ y→0⁺

Famous calculus fact (that could be evaluated also with l'hopital's rule):

lim (sin y)/y = 1

y→0

lim g(y) = 0 • cos 0 • lim y/sin y = 0 • 1 • 1 = 0

y→0⁺ y→0⁺

Finally,

Iim (tan x)^π/2-x = lim f(y) = lim e^g(y) = e⁰ = 1

x→π/2^- y→0⁺ y→0⁺

Answer 2

This problem works only if this is a multiplication and not an exponent.

Then, (π/2 - x) tan x = (π/2 - x)/cot x, which equals 0/0 at x = π/2

Taking the derivative of the numerator and denominator, we get -1/-csc²x

At x = π/2, this equals -1/-csc²x = -1/-(1²) = -1/-1 = 1

By the way, the limit works whether approaching from the left or right side, so this is the limit.