lim(x→0+) (1/x)^(tan x) = ?
i need the limit of:
(1/x)^(tan x)
as x approaches a value just above zero. i'm supposed to use L'Hôspital's rule and i'm stuck at:
e^(lim(x→0+) -(tan x)(ln x))
please explain your answer!
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lim(x→0+) (1/x)^(tan x)
= lim(x→0+) e^[ tanx * ln(1/x) ]
= lim(x→0+) e^[ ln(1/x) / cotx ]
= lim(x→0+) e^[ x * (-1/x^2) / - cosec^2 x ] ... [Using L'Hospital's Rule]
= lim(x→0+) e^[ x * (sinx/x)^2]
= e^[ lim(x→0+) x * lim(x→0+) (sinx/x)^2 ]
= e^0 ... [because lim(x→0+) x = 0 and lim(x→0+) (sinx/x)^2 = 1]
= 1.
I just graphed it on wolframalpha.com and it looks like the lim from 0+ is 1.
ln(L) = lim[x->0⁺] -tan(x)ln(x)
= lim[x->0⁺] -sin(x)ln(x)
= lim[x->0⁺] -ln(x) / csc(x)
=(L'H) lim[x->0⁺] 1 / xcsc(x)cot(x)
= lim[x->0⁺] sin(x)tan(x) / x
= lim[x->0⁺] sin²(x) / x
=(L'H) lim[x->0⁺] 2sin(x)cos(x) = 0
L = exp(0) = 1