In order to find a horizontal asymptote of a function f(x), you need to calculate
[tex] \lim_{x \to \infty} f(x) [/tex] = n
where n is a finite number.
in your case you have to caculate separately the two limits:
a) [tex]\lim_{x \to +\infty} \frac{4 e^{x} + 7 }{ e^{x} - 1 } [/tex] = [tex] \frac{\infty}{\infty} [/tex]
To solve this, you need to regroup eˣ both on the numerator and on the denominator:
[tex]\lim_{x \to +\infty} \frac{4 e^{x} + 7 }{ e^{x} - 1 }[/tex] = [tex]\lim_{x \to +\infty} \frac{4 e^{x} (1 + \frac{7}{ e^{x} } ) }{ e^{x} ( 1 - \frac{1}{ e^{x} } ) }[/tex]
At this point the two eˣ cancel out and the parenthesis tend to 1, therefore
[tex]\lim_{x \to +\infty} \frac{4 e^{x} + 7 }{ e^{x} - 1 }[/tex] = 4
Your first horizontal asymptote is y = 4.
b) Let's calcuate now:
[tex]\lim_{x \to -\infty} \frac{4 e^{x} + 7 }{ e^{x} - 1 }[/tex] = [tex] \lim_{x \to -\infty} \frac{0 + 7}{0 - 1} [/tex] = -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.
