a) One way to approach this part is to set f(x) = 1, and then solve for x. But we won’t get far, because the equation has NO solution. This is because y = 1 is a horizontal asymptote (note that the polynomials in numerator and denominator both have degree 2.)
answer: c = + - inf
b) Note that the graph of f(x) and the graph of g(x) = (x – 2) / (x + 2) are identical, and that they have a removable discontinuity at (-3, 5). Remembering that a point need not always be defined in order for a limit to exist at the point, we see that lim f (x) (as x -> -3) = 5.
answer: c = - 3
c) Rational functions are differentiable, except where the denominator equals 0. There are two possibilities: 1) the removable discontinuity we saw in part (b) or 2) an infinite discontinuity where a vertical asymptote exists. Looking at g(x), defined in part (b), we see a vertical asymptote at x = -2.
answer: c = - 2
