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Question
Evaluate the following limits, if necessary use l’Hôpital Rule:
`lim_(x -> pi/2) (sin x)^tanx`
Solution
`lim_(x -> pi/2) (sin x)^tanx` .....`[1^oo "Indeterminate form"]`
Let g(x) = `(sin x)^tanx`
Taking log on both sides,
log g(x) = tan x log sin x
`lim_(x -> pi/2) log "g"(x) = lim_(x -> pi/2) logsinx/cotx` .....`[0/0 "Indeterminate form"]`
Applying L’ Hôpital’s Rule
= `lim_(x -> pi/2) (cotx/(-"cosec^2x))` = – 1
Exponentiating, we get
`lim_(x -> pi/2) "g"(x) = "e"^-1 = 1/"e"`
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