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lim x->infinity [ (a^x)/{(a^x)+1} ]
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calculus 2?
1
it is also given a>0.
Multiply top and bottom by \[\frac{1}{a^x}\]
This is a rational functions problem you might have seen in precalc. Since the degree of the numerator will always be the same as the degree of the denominator as x --> infinity, the function goes to the ratio of the coefficients of the highest degree term of the top and bottom of the fraction.
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1
which is 1
\[\frac{1}{1+ \frac{1}{a^x}}\]
the second term in bottom goes to zero
we can't say anything about a^x ...because we don't know about "a".(other than only a>0)
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if suppose a=1 then what will happen......i think we will get 1/2
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