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

Answer 1

#y=4# and #y=0# are horizontal asymptotes

Let #y=4/(1+e^(1-x))#
Try to imagine the value of #y# if #x# is made to increase up to #+oo#
As #x# approaches the value #+oo# #y# approaches #4#
Try to imagine the value of #y# if #x# is made to decrease down to #-oo#
As #x# approaches the value #-oo# #y# approaches #0#

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Answer 2

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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