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.
