Practice Problems 5 for 1110 Solution
September 22, 2016
Exercise 1. Find the equation for the asymptote of f (x) =
3x3 −2x2 +x−8
x2 −4
as x tends to infinity.
Solution. Since the highest power of x (the degree) of the numerator is greater than the highest power of x in the denominator
we know that
3x3 − 2x2 + x − 8
=∞
x→∞
x2 − 4
lim
because for large positive x both the numerator and denominator are positive, and
3x3 − 2x2 + x − 8
= −∞
x→∞
x2 − 4
lim
because for large negative values of x the numerator is negative, but the denominator is positive. However, we see that the
degree of the numerator is one more than the degree of the denominator, so there is an oblique asymptote. To figure out what
the asymptote line is we have to do polynomial long division.
3x − 2
x −4
2
3
2
3x − 2x
+x −8
− 3x3
+ 12x
− 2x2 + 13x − 8
2x2
−8
13x − 16
Therefore, our rational function is equal to f (x) = 3x − 2 + 13x−16
x2 −4 . As x gets large both positive or negative, the fraction term
gets arbitrarily small, so for these large values of x the function f (x) gets closer and closer to the line 3x − 2.
Exercise 2. Use continuity to evaluate limx→0 tan(x + sin(x)).
Solution. Since tan(x) is a continuous function, we can move the limit into the argument of tan(x) if the limit on the inside
exists. Since limx→0 x + sin(x) = 0 does indeed exist, limx→0 tan(x + sin(x)) = tan(limx→0 (x + sin(x))) = tan(0) = 0.
Exercise 3. Evaluate limx→∞ sin( π2 e1/x ). Note that sin(x) and ex are both continuous functions.
Solution. Again, sin(t) and et are continuous, so we may move the limit inside.
π
π
π
lim sin( e1/x ) = sin( lim e1/x ) = sin( elimx→∞ 1/x ) = sin(π/2) = 1.
x→∞ 2
2
2
x→∞
We also used the fact that we can move constant around limits.
Exercise 4. What is the domain of f (x) =
continuous at every point?
x2 −x−6
x−3 ?
What is limx→3
x2 −x−6
x−3 ?
How can you alter f (x) to make it defined and
Solution. The domain of f (x) is every real number except 3, i.e. (−∞, 3) ∪ (3, ∞). To find the limit, we see if we can cancel
terms with the numerator.
x2 − x − 6
(x − 3)(x + 2)
= lim
= lim (x + 2) = 5
x→3
x→3
x→3
x−3
x−3
lim
So f (x) has a removable discontinuity at x = 3. If the numerator did not have a factor of (x − 3), then there would be a vertical
asymptote at x = 3. The way to alter f (x) to make it continuous is to define the new function g(x) to be f (x) everywhere except
at x = 3 and to be 5 at x = 3. In fact, this new function is then just g(x) = x + 2.

2

2 + x if x ≤ 0
Exercise 5. Is the following function continuous? If not, where is it discontinuous? f (x) = −2x
if 0 < x ≤ 2

 2
−x
if 2 < x
Solution. For a function to be continuous, it has to be continuous at every point. Since each individual part of the piece-wise
function is continuous, f (x) is continuous on those parts. So all we are left to check are the points at the boundaries of the
pieces, namely x = 0 and x = 2. But we see that the left hand and right hand limits at x = 0 do not coincide.
lim −2x = 0,
x→0+
lim 2 + x2 = 2.
x→0−
 2
x −1

if x < 1
 x−1
Exercise 6. Find the values of a and b that make f (x) continuous everywhere. f (x) = ax2 − bx + 3 if 1 ≤ x < 3


2x − a + b
if 3 ≤ x
Solution. Again, each individual part of the function is continuous, so we only need to check the boundary parts of the domain
pieces, namely x = 1 and x = 3.
lim−
x→1
x2 − 1
(x − 1)(x + 1)
= lim−
= lim− (x + 1) = 2.
x−1
x−1
x→1
x→1
Therefore, we must have
2 = lim ax2 − bx + 3 = a − b + 3 =⇒ b − 1 = a.
x→1+
The limits also have to agree at x = 3:
9a − 3b + 3 = lim− ax2 − bx + 3 = lim+ 2x − a + b = 6 − a + b
x→3
x→3
=⇒ 10a − 2b = 3.
Plugging in the first equation into the second gives us 10(b − 1) − 2b = 3 and solving for b we find b = 13/8 and a = 5/8.
Exercise 7. Use the intermediate value theorem to show that there exists a solution to the equation cos(x) = 2x.
Solution. A solution to the equation cos(x) = 2x is the same as when cos(x) − 2x = 0, or in other words, when the continuous
function f (x) = cos(x) − 2x is equal to 0. By plugging in easy to compute values for x we see that
f (0) = cos(0) − 2(0) = 1
f (π/2) = cos(π/2) − x(π/2) = −π.
Since one value is negative and the other is positive, and the function is continuous, we know by the intermediate value theorem
that the function must be equal to zero somewhere on the interval (−π, 1).
Exercise 8. Use the intermediate value theorem to show that there is a number whose cube is one more than itself.
Solution. For the cube of a number x to be one more than itself, we must have x3 = x + 1, or in other words x3 − x − 1 = 0.
Again, f (x) = x3 − x − 1 is a continuous function because it is a polynomial, and by trial and error we find a point at which the
value of f is positive and a point where it is negative.
f (0) = −1
f (2) = 8 − 2 − 1 = 5.
Therefore, by the intermediate value theorem the function f (x) is zero for some value between 0 and 2. Therefore, there is some
number between 0 and 2 whose cube is one more than itself.