If I define the functions
and
, evaluate
assuming you know the following things about h(x):
- h is continuous everywhere except when
Difficulty:
First of all, we can immediately determine that g(x) contains
the factor (x + 1), because we're told that h must be discontinuous
when 
.
How do we know this? In order for h to be discontinuous when ,
we need the denominator of h, which happens to be g, to
equal 0 when you plug 
in for x.
The second bullet in the problem tells us that the degree of the denominator
must be less than the degree of the numerator, because in order for a rational
function to have an infinite limit at infinity, the denominator degree must
be less than the numerator's. So there can be, at most, two variable factors
in the denominator. We've already got one linear factor in there, so at most
there could be one more. However, you don't have to worry about that! If there
were another linear factor in the denominator, take for example (x
+ k), then h would have to be discontinuous at 
as well, since plugging that value into h would also cause a 0 in
the denominator, and we're told that h is only discontinuous once.
So, what do we know at this point? Not a whole lot, unfortunately. We know
exactly what f is (but we have since the beginning, so that's not
such a big deal), and we almost know what g is. In fact, g
can only differ from (x + 1) by a constant factor, in order for g
to remain equal to 0 if you plug in .
So, let's write out exactly what h equals (by the way, I am going
to go ahead and factor f):
Remember, C is just some unknown real number, since g has
to be some real multiple of (x + 1). Now, let's use the last fact
we are given in the problem (the limit of h as x approaches
) to find C.
Once we find C, we'll know what g and h are, and
we'll be able to answer the question posed by the problem:
So, we know that ,
g(x) = 4(x + 1), and h is equal to f
divided by g. Time to answer the actual question:
