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Math 115- Math 133
Calculus 1
Fall 2022-23 Midterm
24 pts 10 pts 10 pts 24 pts 4 pts 16 pts 12 pts 100 pts
Exam -1
Answer Key
Date: October 30, 2022 Full Name :
Time: 10:00-12:00 Calculus-1 Number :
Student ID :
IMPORTANT: 1. Show all your work to get full credit. 2. Calculators are not allowed.

1 1
1. (2 points each) Given the functions f (x) = x2 + 2x, g(x) = , h(x) = √ ,
x2 + 2x + 3 2 − sin x
find the followings:

(a) Domain(f ) = R (d) Range(f ) = [−1, ∞)

(b) Domain(g) = R (e) Range(g) = (0, 1/2]
√
(c) Domain(h) = R (f) Range(h) = [1/ 3, 1]

2. (2 points each) Write TRUE (T) / FALSE (F) in the corresponding box.

(a) For any function f which is defined on R if f ( n1 ) = 1

n
for any positive integer n, then
lim+ f (x) = 0.
x→0

(b) For any function f which is defined on R if f (x) > x for any x > 100, then
lim f (x) = ∞.
x→∞

(c) For any function f if lim xf (x) = 0, then lim f (x) 6= ∞.

x→0 x→0
0
(d) For any function f which is differentaiable on R if lim f (x) = 1, then lim f (x)−x = 0
x→∞ x→∞
0 00
(e) For any function f if f (x0 ) = 0, then f (x0 ) = 0.
1
(f) For any function f if lim f (x) = −∞, then lim f (x) sin( f (x) ) = 1.
x→1 x→1

a) b) c) d) e) f)
F T F F F T
√ √
x2 x + 2x x + x + 3
3. (10 points) Show that the function f (x) = √ has the oblique
x x+1
asymptote y = x + 2 as x → ∞.

We prove the claim by showing that the following limit is equal to zero.
√ √
x2 x + 2x x + x + 3 1 1
lim f (x)−(x+2) = lim √ −x−2 = lim √ = lim 3/2 =0
x→∞ x→∞ x x+1 x→∞ x x + 1 x→∞ x (1 + x−3/2 )
 1 1
4. (10 points) Show that the equation 2
+ = 0 has a solution in the interval (0, 4).
x (x − 4)3
We put f (x) = x12 + (x−4) 1
3 , then observe that f (1) = 26/27 and f (3) = −8/9. Since f

is continuous on the interval [1, 3], by the IVT we have zero of the function f in the interval
[1, 3], hence in the interval (0, 4).

5. (8 points each) Evaluate the following limits, if they exist, without using the L’Hopital’s
Rule.
x5/2 − 1
(a) lim 3/2
x→1 x −1

x5/2 − 1 x5/2 − 1 (x5/2 + 1)(x3/2 + 1) (x5 − 1)(x3/2 + 1)

lim = lim lim
x→1 x3/2 − 1 x→1 x3/2 − 1 (x5/2 + 1)(x3/2 + 1) x→1 (x3 − 1)(x5/2 + 1)

(x − 1)(1 + x + x2 + x3 + x4 )(x3/2 + 1) (1 + x + x2 + x3 + x4 )(x3/2 + 1)

= lim = lim
x→1 (x − 1)(1 + x + x2 )(x5/2 + 1) x→1 (1 + x + x2 )(x5/2 + 1)
5.2 5
= = .
3.2 3
5x2 + 7x3 sin( x12 ) + 4x4 sin( x12 )
(b) lim
x→∞ 3x2 + x2 cos( √1x ) + 1
 
2
x 5+ 7x sin( x12 )
+ 4x 2
sin( x12 )
5 + 7x sin( x12 ) + 4x2 sin( x12 )
= lim = lim
3 + cos( √1x ) + x−2
 
x→∞ x→∞
2 1 −2
x 3 + cos( x ) + x
√

√
Now, we observe the following calculations: lim cos(1/ x) = 1 and
x→∞

lim x2 sin(1/x2 ) =
x→∞

sin(1/x2 )
 
2 + sin(h)
= lim = We put h = 1/x , h → 0 ⇐⇒ x → ∞ = lim =1
x→∞ 1/x2 h→0+ h
Another limit calculation is the following:

sin(1/x2 )
 
2 +
lim x sin(1/x ) = lim = We put h = 1/x, h → 0 ⇐⇒ x → ∞
x→∞ x→∞ 1/x

sin(h2 ) h sin(h2 )
= lim+ = lim+ = 0.1 = 0
h→0 h h→0 h2
So, we obtain the following:

5 + 7x sin( x12 ) + 4x2 sin( x12 ) 5+0+4 9

lim 1 −2
= =
x→∞ 3 + cos( x ) + x
√ 3+1+0 4
 sin3 (x) sin( x1 )
(c) lim
x→0 x
sin3 (x) sin( x1 ) sin x 2 1
lim = lim sin x sin( ).
x→0 x x→0 x x
We observe that
sin x
lim =1
x→0 x

and for any x ∈ R − {0} we have

1 1
−1 ≤ sin( ) ≤ 1 =⇒ − sin2 x ≤ sin2 x sin( ) ≤ sin2 x.
x x
Since lim ± sin2 x = 0, by the Sandwich Theorem, we get lim sin2 x sin( x1 ) = 0. Hence,
x→0 x→0

sin3 (x) sin( x1 ) sin x 2 1

lim = lim sin x sin( ) = 1.0 = 0.
x→0 x x→0 x x

f (x)
6. (4 points) Find two functions f and g so that lim f (x) − g(x) = ∞ and lim = 1.
x→∞ x→∞ g(x)

f (x) = x2 + x

g(x) = x2
f (x) x2 +x
Clearly lim f (x) − g(x) = lim x2 + x − x2 = ∞ and lim = lim 2 = 1.
x→ ∞ x→ ∞ x→ ∞ g(x) x→ ∞ x

dy
7. (8 points each) Find for each of the followings:
dx
x sin x
(a) y=
x2 + 1
d
(x sin x)(x2 + 1) − (sin x + x cos x)(x2 + 1) − 2x2 sin x
 
dy d x sin x dx
x sin x.2x
= = = .
dx dx x2 + 1 (x2 + 1)2 (x2 + 1)2

(b) y = sec4 (x3 )

dy d
= 4 sec3 (x3 ). (sec(x3 )) = 4 sec3 (x3 ) sec(x3 ). tan(x3 ).3x2 = 12x2 sec4 (x3 ) tan(x3 )
dx dx
8. (12 points) The graph of f (x) = x4 − 32 x + 163
and the graph of g(x) = 12x − 15 in-
tersect at exactly one point and f (x) ≥ g(x) for any x ∈ R. Find the point of intersection
of these graphs.

Regarding to the information given in the quesition, the graphs of f and g have to be tan-
0 0
gent at the point of intersection. So, we solve the equation f (x) = g (x), i.e. 4x3 − 32 = 12
leading to the solution x = 3/2. The point of intersection is, now, (3/2, g(3)) = (3/2, 3).