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Question
Evaluate the following limits, if necessary use l’Hôpital Rule:
`lim_(x -> 0^+) x^x`
Solution
`lim_(x -> 0^+) x^x` .......[0° indeterminate form]
Let g(x) = xx
Taking log on both sides
log g(x) = log xx
log g(x) = x log x
`lim_(x -> 0^+) log "g"(x) = lim_(x -> 0^+) x log x` .......`[0 xx oo "indeterminate form"]`
= `lim_(x -> 0^+) log x/(1/x)` ........`[oo/oo "Indeterminate form"]`
Applying L’ Hôpital’s rule,
= `lim_(x -> 0^+) (1/x)/(- 1/x^2)`
= `lim_(x -> 0^+) - x` = 0
Exponentiating, we get
`lim_(x -> ^+) "g"(x)` = e0
= 1
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