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प्रश्न
Evaluate the following limits, if necessary use l’Hôpital Rule:
`lim_(x -> 0^+) (cos x)^(1/x^2)`
उत्तर
`lim_(x -> 0^+) (cos x)^(1/x^2)` ......`[1^oo "Indeterminate form"]`
let g(x) = `(cos x)^(1/x^2)`
Taking log on both sides,
log g(x) = `log cos x/x^2`
`lim_(x -> 0^+) log "g"(x) = lim_(x -> 0^+) log cos x/x^2` .......`[0/0 "Indeterminate form"]`
Applying L' Hospital’s rule,
= `lim_(x -> 0^+) - tanx/(2x)` ......`[0/0 "Indeterminate form"]`
Again applying L' Hospital’s rule,
= `lim_(x -> 0^+) - (sec^2x)/2 = - 1/2`
Exponentiating, we get
`lim_(x -> 0^+) "g"(x) = "e"^(- 1/2) = 1/sqrt("e")`
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