MATH 126
MIDTERM 1 SOLUTIONS
September 28, 2016
1. (10 pts) Let f (x) be the function whose graph is shown below.
For each problem, give the value, or explain why it does not exist.
(a) lim f (x) = 5
x→1
(b) lim f (x) = 4
x→2
(c) f (2) = 2
(d) lim f (x) does not exist. The reason is that the one-sided limits do not agree:
x→3
lim f (x) = 5, while lim− f (x) = 3.
x→3+
x→3
(e) List all values of x at which f fails to be continuous. Answer: x = −1, 2, 3.
2. (12 pts) Evaluate the following limits.
√
√
1 + 2x − 1
1+2·4−1
3−1
1
=
=
= .
(a) lim
x→4
x
4
4
2
√
√
√
1 + 2x − 1
( 1 + 2x − 1) ( 1 + 2x + 1)
1 + 2x − 1
(b) lim
= lim
· √
= lim √
x→0
x→0
x
x
( 1 + 2x + 1) x→0 x( 1 + 2x + 1)
2
2
2x
√
= lim √
=
= 1.
x→0 x( 1 + 2x + 1)
x→0
1+1
1 + 2x + 1
= lim
−1
. This is a limit of type −1
: since there is a nonzero numerator with denomi0
h→1 h − 1
nator zero, the limit will equal ±∞. When h < 1 is very close to 1, the expression looks
−1
−1
like (tiny negative
, which is a large positive number. Therefore lim−
= ∞.
number)
h→1 h − 1
(c) lim−
sin(x)
= 1. (This is one of the main results of §3.4.)
x→0
x
(d) lim
3. (20 pts) Use the definition of derivative to compute the derivative of f (x) = x2 + 5.
f (x + h) − f (x)
(x + h)2 + 5 − (x2 + 5)
= lim
h→0
h→0
h
h
f 0 (x) = lim
2xh + h2
h(2x + h)
x2 + 2xh + h2 − x2
= lim
= lim
= lim
h→0
h→0
h→0
h
h
h
= lim (2x + h) = 2x.
h→0
We can check that this answer is correct by using the power rule:
4. (24 pts) Shown below is the graph of the equation y =
d
(x2
dx
+ 5) = 2x.
2ex
.
x2 + 1
5
2.5
-2.5
(a) (10 pts) Find the derivative
dy
.
dx
0
2.5
5
By the quotient rule,
(x2 + 1)2ex − 2ex (2x)
dy
=
.
dx
(x2 + 1)2
(b) (10 pts) Sketch the tangent line at the y-intercept on the above graph, and find an equation
for it.
0
Using the derivative computed above, the slope of the tangent line is f 0 (0) = 2e
= 2.
12
The height of the graph at the y-intercept is f (0) = 21 = 2. So the equation for the
tangent line is
y = 2x + 2.
(c) (4 pts) One point on the curve has a horizontal tangent line. Find its x-coordinate.
A horizontal tangent line occurs when the derivative is equal to 0. Simplifying the answer
in part (a), we set
dy
2ex (x2 + 1 − 2x)
2ex (x2 − 2x + 1)
=
=
= 0.
dx
(x2 + 1)2
(x2 + 1)2
A fraction equals zero only when its numerator equals zero. Since 2ex is never zero, this
happens when x2 − 2x + 1 = (x − 1)2 = 0, i.e. when x = 1.
5. (10 pts) Find all horizontal asymptotes for the graph of the function
f (x) =
x2 − 4
.
2x − x2
Justify your answer by computing the necessary limits.
Horizontal tangent lines arise when a function has a finite limit as x → ±∞.
x2
x2 − 4
=
lim
= −1.
x→∞ −x2
x→∞ 2x − x2
lim
So y = −1 is a horizontal asymptote. Notice that the function has the same limit as
x → −∞, so there is only one horizontal asymptote.
6. (8 pts) Using the Intermediate Value Theorem, explain why the function f (x) = 2x3 − 1
has a zero between x = 0 and x = 1.
We observe that f (0) = −1 and f (1) = 1. Since f is continuous (like all polynomials), and
0 is a y-value between f (0) and f (1), the intermediate value theorem says that there exists
c between 0 and 1 such that f (c) = 0.
7. (16 pts) A bug crawls along the number line. Her position at time t (in seconds) is
given by s(t) = t2 et meters.
(a) Find her average velocity between times t = 0 and t = 5.
vave =
s(5) − s(0)
25e5 − 0
=
= 5e5 m/sec.
5−0
5
(b) Find her (instantaneous) velocity at time t = 1. Notice that
v(t) = s0 (t) = t2 et + 2tet
by the product rule. So
v(1) = e + 2e = 3e m/sec.