DmbAs mf can learnnn idiottttttttt
has two points of discontinuity at
, where
.
We have

so as long as
, we can cancel
and be left with
. Then as
, we have

The limit exists, so the discontinuity at -1 is removable (so definitely not C or D).
Meanwhile,

does not exist, so the discontinuity at 1 is non-removable, making A the correct answer.
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1.
has a horizontal asymptote at 
This means that

(for at least one of these limits)
2.
has a vertical asymptote at 
This means that
has a non-removable discontinuity at
. Since
is some rational function, there must be a factor of
in its denominator.
3.
has an
-intercept at (1, 0)
This means
.
(a) With

the second point above suggests
. The first point tells us that

In order for the limit to be 0, the denominator's degree should exceed the numerator's degree; the only way for this to happen is if
so that the linear terms vanish.
The third point tells us that

So

(b) Since

we find that
, and
and
.