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Question

Suppose that a and b (ba) are real positive number the value of limt0(b1+ta1+tba)1/t has the value equals to

A
alnbblnaba
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B
blnbalnaba
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C
blnbalna
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D
(bbaa)(1/(ba))
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Solution

The correct option is D (bbaa)(1/(ba))
Given limt0(b1+ta1+tba)1tab

If L=limxaf(x)g(x)

limxaf(x)1andlimxag(x)

L=elimxa(f(x)1).g(x)
Thus,
L=limt0eb1+ta1+tba1.1t

=e1(ba)limt01t{bt+1at+1}(ba)(00) form
Using L'Hospital's Rule

=e1(ba)limt0[bt+1btat+1at]

e1(ba)limt0[bt+1lnbat+1lna]=eblnbalnaba=e⎜ ⎜ ⎜ ⎜ ⎜lnbbbalnaaba⎟ ⎟ ⎟ ⎟ ⎟

elnbbaa1(ba)=(bbaa)1(ba)

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