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प्रश्न
Evaluate the following limits:
`lim_(x -> oo) x [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)]`
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उत्तर
We know `lim_(x -> 0) ("e"^x - 1)/x` = 1
`lim_(x -> 0) ("a"^x - 1)/x` = log a
`lim_(x -> 0) (1 - cosx)/x` = 0
`lim_(x -> oo) [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)] = lim_(x -> oo) [(3^(1/x) + 1 - cos(1/x) - "e"^(1/x))/(1/x)]`
= `lim_(x -> oo) [(3^(1/x) - 1 + 1 - "e"^(1/x))/(1/x) + (1 - cos(1/x))/(1/x)]`
= `lim_(x -> oo) [((3^(1/x) - 1) - ("e"^(1/x) - 1))/(1/x) + (1 - cos(1/x))/(1/x)]`
= `lim_(x > 0)[(3^(1/x) - 1)/(1/x) - ("e"^(1/x) - 1)/(1/x) + (1 -cos(1/x))/(1/x)]`
Put y = `1/x`
When x = `oo`
⇒ y = `1/oo` = 0
`lim_(x -> oo) x [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)] = lim_(y - 0) [(3y - 1)/y - ("e"^y - 1)/y + (1 - cosy)/y]`
= `(lim_(y -> 0) (3^y - 1)/y) -(lim_(y -> 0) ("e"^y - 1)/y) + (lim_(y -> 0) (1 - cosy)/y)`
= `log 3 - 1 + 0`
`lim_(x -> oo) x [3^(1/x) + 1 - cos(1/x) - "e"^(1/x)] = (log 3) - 1`
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