Math 151.02 - Midterm 1 Solutions
(1) Find the derivative of the function
f (x) =
1
.
x+1
Using the definition of the derivative f 0 (x), we have
f (x + h) − f (x)
h→0
h
1
− 1
= lim x+h+1 x+1
h→0
h
x+1
x+h+1
(x+h+1)(x+1) − (x+1)(x+h+1)
= lim
h→0
h
x + 1 − (x + h + 1)
= lim
h→0 h(x + h + 1)(x + 1)
−h
= lim
h→0 h(x + h + 1)(x + 1)
−1
= lim
h→0 (x + h + 1)(x + 1)
−1
=
.
(x + 1)2
f 0 (x) = lim
2
(2) Using the graph below, determine (approximately) the value of the following limits, or
explain why the limit does not exist.
(a) lim f (x)
x→1
Since the function is continuous at 1,
lim f (x) = f (1) = 2.
x→1
(b) lim f (x)
x→∞
Since the graph keeps going between −1 and 1 as x goes to ∞, there is no limit.
(c) lim
h→0
f (4 + h) − f (4)
h
This limit is the definition of f 0 (4). But f 0 (4) is the slope of the tangent line at 4,
which we can estimate with the graph. It looks like it’s about −1.
(d) f 0 (5)
Again, f 0 (5) is the slope of the tangent line at x = 5. But the graph has a corner at
x = 5, so f 0 (5) does not exist.
Math 151.02
3
(3) (a) On the graph of g(x) below, pick an interval (1 − ε, 1 + ε) such that for no interval
(2 − δ, 2 + δ) is it true that if x is in (2 − δ, 2 + δ), then g(x) is in (1 − ε, 1 + ε). Briefly
explain (or indicate on the diagram) why your choice of interval works.
Note that no matter how small δ is, numbers just to the left of x = 2 get sent a long
way from y = 1. So it is not true that if x is in (2−δ, 2+δ), then g(x) is in (1−ε, 1+ε).
(b) What does the ε you found in part (a) tell you about lim g(x)?
x→2
This tells us that the limit cannot be 1. (If the limit were 1, then for every ε > 0 we
could find a δ > 0. But in part (a), we found an ε for which there is no δ.)
4
(4) Calculate the following limits:
(a) lim x csc x
x→0
lim x csc x = lim x ·
x→0
x→0
= lim
x→0
1
sin x
1
sin x
x
limx→0 1
limx→0 sinx x
1
=
1
= 1.
=
(b) lim
x→π
tan(x) + 1
.
x
Since tan x and x are continuous at x = π, and we have sums and quotients of continuous functions. So (tan(x) + 1)/x is continuous at π, so we can just plug in π to
get
tan(x) + 1
tan(π) + 1
1
lim
=
= .
x→π
x
π
π
Math 151.02
5
(5) For which values of a is the function
(
2ax2 − a2 , x < 1
f (x) =
4a − x,
x≥1
continuous for all x?
No matter what a is, if x > 1, then f (x) = 4a − x. Since this is continuous, f is
continuous for x > 1. Similarly, for if x < 1, then f (x) = 2ax2 − a2 . This is continuous no
matter what a is, so f is continuous for x < 1. The only question is when f is continuous
at x = 1.
On one hand, we have
lim f (x) = f (1) = 4a − 1,
x→1+
and on the other hand, we have
lim f (x) = 2a − a2 .
x→1−
For f to be continuous at 1, these two limits must be the same. So solve
2a − a2 = 4a − 1
0 = a2 + 2a − 1
p
−2 ± 4 − 4(1)(−4)
a=
√ 2
= −1 ± 2.
√
√
So f (x) is continuous for all x only when a is −1 − 2 or −1 + 2.
6
(6) If f is a function whose domain is all real numbers, and
2x < f (x) < 4x
for all x > 0, what is the value of lim f (x)? Briefly explain your answer.
x→0+
Since
lim 2x = lim 4x = 1,
x→0+
x→0+
the sandwich theorem says that
lim f (x) = 1.
x→0+