Evaluate the limit \lim_{x\rightarrow 0^+}(1/tan(x))^{1/ln(x)}?

1 Answer
Feb 11, 2018

1/e

Explanation:

let y=lim_(xrarr0^+)((1/tan(x))^(1/ln(x)))

take ln() of both sides: lny=ln[lim_(xrarr0^+)((1/tan(x))^(1/ln(x)))]

move ln() into the limit: lny=lim_(xrarr0^+)(ln[(1/tan(x))^(1/ln(x))])

logarithm properties: lny=lim_(xrarr0^+)((1/ln(x))ln(1/tan(x)))
lny=lim_(xrarr0^+)(ln(1/tan(x))/ln(x))

if you plug in x=0^+ directly, you get: ln(1/tan(0^+))/ln(0^+)
=ln(1/0^+)/ln(0^+)
=ln(oo)/-oo
=oo/-oo
indeterminate, so use l'hopital's rule:

lny=lim_(xrarr0^+)((d/dx[ln(1/tan(x))])/(d/dx[ln(x)]))
lny=lim_(xrarr0^+)((tan(x)*d/dx(1/tan(x)))/(1/x))
lny=lim_(xrarr0^+)(xtan(x)*d/dx(cot(x)))
lny=lim_(xrarr0^+)(xtan(x)*(-csc^2(x)))
lny=lim_(xrarr0^+)((-xsin(x)/cos(x))/sin^2(x))
lny=lim_(xrarr0^+)(-x/(cos(x)sin(x)))

again, plugging in x=0^+ results in 0/0 (indeterminate), so use l'hopitals rule again:
lny=lim_(xrarr0^+)((d/dx[-x])/(d/dx[(cos(x)sin(x))]))
lny=lim_(xrarr0^+)(-1/(-sin^2(x)+cos^2(x)))

now plug in x=0^+:
lny=lim_(xrarr0^+)(-1/(-sin^2(0^+)+cos^2(0^+)))
lny=lim_(xrarr0^+)(-1/1)
lny=-1
y=e^(-1)
y=1/e

you've solved for y, which equals the limit you want to find.

check with this graph: graph{(1/tan(x))^(1/ln(x)) [-1.695, 2.15, -0.3, 1.742]}

as x approaches 0 from the right, the limit is about 0.37, or 1/e