Question

What are the equations of the horizontal asymptotes of y = (4e^x+7)/(e^x-1) in the xy plane?

Answer 1

In order to find a horizontal asymptote of a function f(x), you need to calculate

`\lim_(x \to \infty) f(x)` = n
where n is a finite number.

in your case you have to caculate separately the two limits:

a)
`\lim_(x \to +\infty) (4 e^(x) + 7 )/( e^(x) - 1 )` =
`(\infty)/(\infty)`

To solve this, you need to regroup eˣ both on the numerator and on the denominator:

`\lim_(x \to +\infty) (4 e^(x) + 7 )/( e^(x) - 1 )` =
`\lim_(x \to +\infty) (4 e^(x) (1 + (7)/( e^(x) ) ) )/( e^(x) ( 1 - (1)/( e^(x) ) ) )`

At this point the two eˣ cancel out and the parenthesis tend to 1, therefore

`\lim_(x \to +\infty) (4 e^(x) + 7 )/( e^(x) - 1 )` = 4
Your first horizontal asymptote is y = 4.

b) Let's calcuate now:


`\lim_(x \to -\infty) (4 e^(x) + 7 )/( e^(x) - 1 )` =
`\lim_(x \to -\infty) (0 + 7)/(0 - 1)` = -7
Therefore your second horizontal asymptote is y = -7

Your answer is, hence, y = 4 and y = -7 are horizontal asymptotes of the given function.