We get 0^0, so it's indeterminate
L = sin(x)^tan(x)
ln(L) = tan(x) * ln(sin(x))
ln(L) = ln(sin(x)) / cot(x)
f(x) = ln(sin(x))
f'(x) = cos(x) / sin(x) = cot(x)
g(x) = cot(x)
g'(x) = -csc(x)^2
ln(L) => -cot(x) / csc(x)^2
ln(L) => -cos(x) / (sin(x) * 1/sin(x)^2)
ln(L) => -cos(x) / (1/sin(x))
ln(L) => -sin(x)cos(x)
ln(L) => (-1/2) * sin(2x)
x goes to 0
ln(L) => (-1/2) * sin(0) = (-1/2) * 0 = 0
ln(L) = 0
L = e^0
L = 1
The limit is 1
sinh(x) / cosh(x) =>
((1/2) * (e^(x) - e^(-x))) / ((1/2) * (e^(x) + e^(-x))) =>
(e^(x) - e^(-x)) / (e^(x) + e^(-x)) =>
e^(-x) * (e^(2x) - 1) / (e^(-x) * (e^(2x) + 1)) =>
(e^(2x) - 1) / (e^(2x) + 1) =>
(e^(2x) + 1 - 2) / (e^(2x) + 1) =>
(e^(2x) + 1) / (e^(2x) + 1) - 2 / (e^(2x) + 1) =>
1 - 2/(e^(2x) + 1)
x goes to infinity
1 - 2/(e^inf + 1)
1 - 2/inf
1 - 0
1
1)
Let y= (sin x)^tan x
ln y = tan x ln (sin x)
lim x-->0+ ln y = lim x-->0+ tanx ln (sin x)
= lim x-->0+ ln (sin x) /(1/tan(x))
= lim x-->0+ ln (sin x) / cot x
The numerator and denominator approach ∞inity so we can apply L'Hopital's rule
= lim x-->0+ (cos x /sin x) / (-csc^2 x)
= lim x-->0+ -cot x / csc^2 x
= lim x-->0+ csc^2 x / (-2 csc x csc x cot x)
= lim x-->0+ -1 /(2 cot x)
= lim x-->0+ (-tan x) /2
= 0
lim x-->0+ ln (y) = 0
lim x-->0+ y = e^0 = 1
2)
lim x-->∞ sinh x / cosh x
= lim x-->∞ (e^(x)-e^(-x)) /(e^x+e^(-x))
∞/∞ form
Apply L'Hopital's rule
= lim x-->∞ ( e^(2x) -1) /(e^(2x)+1)
= lim x-->∞ 2 e^(2x) /(2e^(2x))
= lim x-->∞ 1
= 1
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Answers & Comments
Verified answer
We get 0^0, so it's indeterminate
L = sin(x)^tan(x)
ln(L) = tan(x) * ln(sin(x))
ln(L) = ln(sin(x)) / cot(x)
f(x) = ln(sin(x))
f'(x) = cos(x) / sin(x) = cot(x)
g(x) = cot(x)
g'(x) = -csc(x)^2
ln(L) => -cot(x) / csc(x)^2
ln(L) => -cos(x) / (sin(x) * 1/sin(x)^2)
ln(L) => -cos(x) / (1/sin(x))
ln(L) => -sin(x)cos(x)
ln(L) => (-1/2) * sin(2x)
x goes to 0
ln(L) => (-1/2) * sin(0) = (-1/2) * 0 = 0
ln(L) = 0
L = e^0
L = 1
The limit is 1
sinh(x) / cosh(x) =>
((1/2) * (e^(x) - e^(-x))) / ((1/2) * (e^(x) + e^(-x))) =>
(e^(x) - e^(-x)) / (e^(x) + e^(-x)) =>
e^(-x) * (e^(2x) - 1) / (e^(-x) * (e^(2x) + 1)) =>
(e^(2x) - 1) / (e^(2x) + 1) =>
(e^(2x) + 1 - 2) / (e^(2x) + 1) =>
(e^(2x) + 1) / (e^(2x) + 1) - 2 / (e^(2x) + 1) =>
1 - 2/(e^(2x) + 1)
x goes to infinity
1 - 2/(e^inf + 1)
1 - 2/inf
1 - 0
1
1)
Let y= (sin x)^tan x
ln y = tan x ln (sin x)
lim x-->0+ ln y = lim x-->0+ tanx ln (sin x)
= lim x-->0+ ln (sin x) /(1/tan(x))
= lim x-->0+ ln (sin x) / cot x
The numerator and denominator approach ∞inity so we can apply L'Hopital's rule
= lim x-->0+ (cos x /sin x) / (-csc^2 x)
= lim x-->0+ -cot x / csc^2 x
= lim x-->0+ csc^2 x / (-2 csc x csc x cot x)
= lim x-->0+ -1 /(2 cot x)
= lim x-->0+ (-tan x) /2
= 0
lim x-->0+ ln (y) = 0
lim x-->0+ y = e^0 = 1
The limit is 1
2)
lim x-->∞ sinh x / cosh x
= lim x-->∞ (e^(x)-e^(-x)) /(e^x+e^(-x))
∞/∞ form
Apply L'Hopital's rule
= lim x-->∞ ( e^(2x) -1) /(e^(2x)+1)
= lim x-->∞ 2 e^(2x) /(2e^(2x))
= lim x-->∞ 1
= 1