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\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\to -infty} e^{x-2}=lim_{x\to -infty} e^x / e^2= 1/ e^2lim_{x\to -infty} e^x=1/ e^2 * 0 = 0#

- Changing variables, you can call #y=x-2#, and note that if #x->-infty#, so does #y#. So,

#lim_{x\to -infty} e^{x-2}=lim_{y\to -infty} e^y =0#.

**Positive infinity** #lim_{x\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-> infty#. The test to check for oblique ones requires to check the limit of #f(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->infty#.