# How do you find the horizontal asymptote for #4/(1+e^(1-x)) #?

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To find the horizontal asymptote of ( \frac{4}{1+e^{1-x}} ), we examine the behavior of the function as ( x ) approaches positive or negative infinity. Since the exponential term in the denominator has a base of ( e ) and is raised to a negative power as ( x ) approaches positive infinity, the exponential term tends towards 0. Therefore, the horizontal asymptote is ( y = \frac{4}{1+0} = 4 ).

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