Nudar H.

asked • 10/27/21# Evaluate the limit using l'hopital's rule

Iim (tan*x*)^{π/2-x}

*x*→*π/*2^{- }

- Lim x tends to π/2 (negative)
- (tanx)
^{π/2-x}

## 2 Answers By Expert Tutors

Vitaliy V. answered • 10/28/21

Math and Statistics Tutor with 30+ Years of Teaching Experience

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/y^{2}

So, lim g(y) = lim ((cos y / sin y) / 1/y^{2}) = 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) } = e^{0 }= 1

^{ }*x*→*π/*2^{- }y→0^{+} y→0^{+}

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/-csc^{2}x

At x = π/2, this equals -1/-csc^{2}x = -1/-(1^{2}) = -1/-1 = 1

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

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Joel L.

10/27/21