How do you find the horizontal asymptote for g(x)=e^x-2?

Jan 14, 2016

Evaluate x at both $\pm \infty$

Explanation:

$\text{_(xrarroo)^"lim} \left({e}^{x} - 2\right) = \infty$

So, the right limit, as $x \rightarrow \infty$, has no horizontal asymptote.

However, the left limit does have an asymptote:

$\text{_(xrarr-oo)^"lim} \left({e}^{x} - 2\right) = 0 - 2 = - 2$

So, the left limit, as $x \rightarrow - \infty$, has horizontal asymptote $= - 2$.

hope that helped