4. (21 pts) (I) (6 pts) Determine the value of a constant b for which
the function g is continuous at 0.
Show your work. Justify your answer!
2 x +b
if x<0,
X-5
g(x)
=
if x 20, x#4.
We have to show that lim g (x) = g (0).
g(0) = 0 +18 = -1;
lim
X0*
0 + 16
) = lim x + 16
**0* x2 - 160 - 16-
Since lim g (x) has to exist (in order for g to be continuous at x = 0),
it has to be true that
x-0-
x0+
X-
0
lim g(x) - lim g(x) = lim g(x).
Therefore,
-1, and , so b = 5.
so, if b = 5, lim g (x) = g(0),
and therefore, g is continuous at x = 0.
(II) (4 pts) Evaluate the limit. Show your work!
2. b
limx-
(a) lim
g(x) = lim
*+-
2 x + b
X - 5
:
1
* +16
10 g (x) = lim
x++
- Elim
x2 - 16 x++
(x - 10).
OR
2x+5. 2;
(a) lim g (x) = lim
x--00
X-
(b) lim
x++
2 + 16
g(x) = lim
**** x2 - 16
X-+
(III) (4 pts) Determine whether g has any horizontal asymptotes. If so,
write the equation (or equations) of all horizontal asymptotes.
y = 2, by part (II) (a)
y = 0, by part (II) (b)
(IV) (7 pts) Determine whether g has any vertical asymptotes. If so,
write the equation (or equations) of all vertical asymptotes.
Show your work. Justify your answer!
5 + 16
There is no vertical asymptote at x = 5, because lim g(x) =
x+5
52-16'
-8
+ b
(x)
=
there is no vertical asymptote at x = -4, because lim
X -4
- 4
- 5
But,
lim g(x) = lim + 16
x+4+ x2 - 16
lim_ (x + 16) -20
x+4+ (x - 4) 60 (x + 4) +8
(or the numerator goes to 20, and the denominator
is positive and goes to o, as x + 4*)
=
+00
So, the function g has only one vertical asymptote,
x = 4.