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Given a real, f(x)=ln((2+x)/(2-x)),calculate limit lim x--> infinite (x^a*f(1/x))=?  

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lixalixa | (Level 2) Honors

Posted May 24, 2013 at 5:03 PM via web

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Given a real, f(x)=ln((2+x)/(2-x)),calculate limit lim x--> infinite (x^a*f(1/x))=?

 

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 24, 2013 at 5:32 PM (Answer #1)

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You need to evaluate the following limit, such that:

`lim_(x->oo) x^a*f(1/x)`

Replacing `1/x` for `x` in the equation of function yields:

`f(1/x) = ln((2 + 1/x)/(2 - 1/x)) => f(1/x) = ln((2x + 1)/(2x - 1)) `

Replacing `ln((2x + 1)/(2x - 1))` for `f(1/x)` in limit yields:

`lim_(x->oo) x^a*ln((2x + 1)/(2x - 1))`

Using the properties of logarithms yields:

`lim_(x->oo) ln((2x + 1)/(2x - 1))^(x^a) = lnlim_(x->oo) ((2x + 1)/(2x - 1))^(x^a)`

`lnlim_(x->oo) ((2x + 1)/(2x - 1))^(x^a) = lnlim_(x->oo) ((2x + 1)/(2x - 1))^(lim_(x->oo)(x^a)) = 1^oo`

The indetermination of type `1^oo` requests for you to use the  following special limit `lim_(x->oo)(1 + 1/x)^x = e` .

You need to start by adding to and subtracting 1 from `(2x + 1)/(2x - 1)` such that:

`lnlim_(x->oo) (1 + (2x + 1)/(2x - 1) - 1)^(x^a)`

`lnlim_(x->oo) ((1 + 2/(2x - 1))^((2x-1)/2))^((2x^a)/(2x-1)) = ln e^lim_(x->oo)((2x^a)/(2x-1))`

Using the properties of logarithms yields:

`ln e^lim_(x->oo)((2x^a)/(2x-1)) = lim_(x->oo)((2x^a)/(2x-1)) *ln e`

Since `ln e = 1` yields:

`ln e^lim_(x->oo)((2x^a)/(2x-1)) = lim_(x->oo)((2x^a)/(2x-1))`

You need to evaluate the limit in the following cases, such that:

`a > 1 => lim_(x->oo)((2x^a)/(2x-1)) = oo`

`a = 1 => lim_(x->oo)((2x^a)/(2x-1)) = 1`

`a < 1=> lim_(x->oo)((2x^a)/(2x-1)) = 0`

Hence, evaluating the given limit, considering three possiblities for a, yields` lim_(x->oo)((2x^a)/(2x-1)) = oo (a>1), lim_(x->oo)((2x^a)/(2x-1)) = 1 (a = 1), lim_(x->oo)((2x^a)/(2x-1)) = 0 (a < 1).`

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