# How do you find all the asymptotes for function g(x)=e^(x-2)?

Oct 26, 2015

The only asymptote is horizontal, and it is $y = 0$, as $x \to - \infty$.

#### Explanation:

A function can have horizontal asymptotes (if the limits as $x$ approaches $\pm \infty$ exist and are finite), vertical ones (if the domain is not the whole real number set), or oblique ones (if the end behaviour is asymptotically equivalent to a line. Note that, for each of the two directions towards infinity, horizontal and oblique asymptotes are mutually exclusive (actually, horizontal ones are a special case of the oblique ones, since they represent a line with zero slope).

Since the exponential is defined over the whole real set, you can't have vertical asymptotes. So, let's check the infinite limits:

Negative infinity: ${\lim}_{x \setminus \to - \infty} {e}^{x - 2}$

This limit is very similar to the one of ${e}^{x}$, and you can come back to it in two ways:

• Using the fact that ${e}^{x - 2} = {e}^{x} / {e}^{2}$, you have that

${\lim}_{x \setminus \to - \infty} {e}^{x - 2} = {\lim}_{x \setminus \to - \infty} {e}^{x} / {e}^{2} = \frac{1}{e} ^ 2 {\lim}_{x \setminus \to - \infty} {e}^{x} = \frac{1}{e} ^ 2 \cdot 0 = 0$

• Changing variables, you can call $y = x - 2$, and note that if $x \to - \infty$, so does $y$. So,

${\lim}_{x \setminus \to - \infty} {e}^{x - 2} = {\lim}_{y \setminus \to - \infty} {e}^{y} = 0$.

Positive infinity ${\lim}_{x \setminus \to \infty} {e}^{x - 2}$

Everything we said above, still holds, so you find that the limit is $+ \infty$, either by algebra (${e}^{x - 2} = {e}^{x} / {e}^{2}$), or by substitution ($y = x - 2$). So, there is no horizontal asymptote as $x \to \infty$. The test to check for oblique ones requires to check the limit of $f \frac{x}{x}$. Is that limit is finite, it represents the slope of the line. But I'm assuming that you know that if you divide an exponential by a polynomial (of any degree), the exponential always "wins", and so the limit is still infinite, and there is no oblique asymptote when $x \to \infty$.