第一次期中考

1. Student ID: U10011024 Name: Kuan-Lun Wang
√
1. Let f (x) = 5 x − x3.
(a) Give the domain and range of f .
(b) State whether f is odd, even or neight.
(a) Range of f is R because that y 5 = x − x3. Domain of f
is R because taht x3 − x + y 5 = 0 and range of f is R.
(b) f is odd because that f (−x) = −f (x).
2. Find that following linits.
tan x
(a) lim .
x→0 x
[x2] − x2
(b) lim where [·] is the Gaussian symbol.
x→2+ x − 2
1
x2 sin( x )
(c) lim .
x→0 sin x
tan x sin x 1
(a) lim = lim lim = 1 · 1 = 1.
x→0 x x→0 x x→0 cos x
[x2] − x2 4 − x2
(b) lim = lim = lim −(x + 2) = −4.
x→2+ x − 2 x→2+ x − 2 x→22
1
1 x2 sin( x )
(c) Note that −1 ≤ lim sin( ) ≤ 1. We have lim =
x→0 x x→0 sin x
x 1 1
lim lim x lim sin( ) = 1 · 0 · lim sin( ) = 0.
x→0 sin x x→0 x→0 x x→0 x
3. Find a simplify expression for the product
1 1 1
(1 − )(1 − ) · · · (1 − )
2 3 n
and verify its validity for all integers n ≥ 2.
1
We can show by mathematical induction that (1 − 2 )(1 −
1 1 1
3 ) · · · (1 − n ) = n for all integers n ≥ 2. The basis step,
1 1
namely, the case where n = 2, hold because (1 − 2 ) = 2 .
Calculus First Test 2011/10/20 1

2. Student ID: U10011024 Name: Kuan-Lun Wang
Now, assume that (1 − 1 )(1 − 1 ) · · · (1 − n ) = n ; this is the
2 3
1 1
inductive hypothesis. To complete the proof, we must show,
under the assumption that the inductive hypothesis is true,
1
that (1 − 2 )(1 − 1 ) · · · (1 − n+1 ) = n+1 . Using the inductive
3
1 1
hypothesis, we have
1 1 1
(1 − )(1 − ) · · · (1 − )
2 3 n+1
1 1 1 1
=((1 − )(1 − ) · · · (1 − ))(1 − )
2 3 n n+1
1 n 1
= · = .
n n+1 n+1
This completes both the inductive step and the proof.
√ √
4. Given an , δ proof of lim x = c for c>0.
x→c
√
Give >0, let δ = c . Then if |x − c|<δ, we have
√
√ √ |x − c| δ c
| x − c| = √ √ <√ √ =√ √ ≤ .
x+ c x+ c x+ c
This establishes the limit statement using the definition.
f (x)
5. Prove that if lim [ ] exists, then lim f (x) = 0.
x→0 x x→0
f (x) f (x)
If lim [ ] exists, then lim f (x) = lim [ ] lim x =
x→0 x x→0 x→0 x x→0
f (x)
lim [ ] · 0 = 0.
x→0 x
2
6. Express dxy in terms of x and y for b2x2 + a2y 2 = a2b2
d
2
where a, b are positive constants.
Calculus First Test 2011/10/20 2

3. Student ID: U10011024 Name: Kuan-Lun Wang
dy 2
Note that dx = − a2x because that b2x2 + a2y 2 = a2b2. We
b
y
have
dy
d2y d b2 x b2(a2y) − b2x(a2 dx )
= (− )=−
dx2 dx a2y a4 y 2
b2
a2b2y + a2b2x a2x y a4 b2 y 2 + a2 b4 x 2
=− =−
a4 y 2 a6 y 3
a2b2(b2x2 + a2y 2) a2b2(a2b2)
=− =−
a6y 3 a6y 3
b4
= − 2 3.
ay
7. Let f and g be continuous functions such that f (0) < g(0)
< g(1) < f (1). Prove that there exists c ∈ (0, 1) such that
f (c) = g(c).
Let h(x) = f (x)−g(x). This is a continuous function of h
because that f and g are continuous functions. We see h is a
differentiable function where x ∈ (0, 1) because it continuous
where x ∈ [0, 1]. Note that h(0)<0 and h(1)>0. It is that
exists c ∈ (0, 1) such that f (c) = g(c).
x sin(1/x) x = 0
8. Let f (x) = and g(x) = xf (x).
0 x=0
(a) Show that f and g are both continuous at 0.
(b) Show that f is not differentiable at 0.
(c) Show that g is differentiable at 0 and give g (0).
(a) lim f (x) = lim x sin(1/x) = 0 = f (0) and lim g(x) =
x→0 x→0 x→0
lim xf (x) = 0 = g(0).
x→0
Calculus First Test 2011/10/20 3

4. Student ID: U10011024 Name: Kuan-Lun Wang
f (x) − f (0)
(b) f (0) = lim = lim sin(1/x) is not differen-
x→0 x−0 x→0
tiable at 0.
1 g(x) − g(0)
(c) Note that −1 ≤ lim sin( ) ≤ 1. g (0) = lim =
x→0 x x→0 x−0
x2 sin(1/x)
lim = lim x sin(1/x) = 0 is differentiable at 0.
x→0 x−0 x→0
9. Find the indicated derivatives.
d2
(a) dx2 [(x2 − 3x) dx (x + x−1)].
d
d d2
(b) dt [t2 dt2 (t cos 3t)].
d2 −1 d2
(a) dx
2 d
2 [(x − 3x) dx (x + x )] = dx2 [(x2 − 3x)(1 − x−2)] =
d2
dx2
[x2 − 3x − 1 + 3x−1] = dx [2x − 3 − 3x−2] = 2 + 6x−3.
d
d2 d
(b) Note that dt2 (t cos 3t) = dt (cos 3t−3t sin 3t) = −6 sin 3t−
d d
9t cos 3t, dt (t2 sin 3t) = 2t sin 3t+3t2 cos 3t and dt (t3 cos 3t) =
2 3
3t cos 3t − 3t sin 3t. We havt
d 2 d2
[t (t cos 3t)]
dt dt2
d
= [t2(−6 sin 3t − 9t cos 3t)]
dt
d
= − 3 [2t2 sin 3t + 3t3 cos 3t]
dt
= − 3[(4t sin 3t + 6t2 cos 3t) + (9t2 cos 3t − 9t3 sin 3t)]
=27t3 sin 3t − 12 sin 3t − 45 cos 3t.
10. Find the points (c, f (c)) where the line tangent to the
x
graph of f (x) = x+1 is parallel to the secent line that passes
through the points (1, f (1)) and (3, f (3)).
1 1(x+1)−x(1)
Note that f (c) = c2 +2c+1
because tat f (x) = (x+1)2
=
Calculus First Test 2011/10/20 4

5. Student ID: U10011024 Name: Kuan-Lun Wang
1
x2 +2x+1
. The points (c, f (c)) where the line tangent to the
x c 1
graph of f (x) = x+1 is y − c+1 = c2+2c+1 (x − c). Note that
f (3)−f (1) 1
√ 1 1
= 8 . c = ± 2 − 1 because that c2+2c+1 = 8 . This
3−1 √ √
is that the points (± 2 − 1, f (± 2 − 1)) where the line
√
± √2−1
√
√1
tangent y − ± 2 = (± 2−1)2 (x − ± 2 + 1) to the graph
x
of f (x) = x+1 parallel to the secent line that passes through
the points (1, f (1)) and (3, f (3))
11. Prove or give a counterexample:
(a) Suppose that f (x)<g(x) for all x ∈ (c − p, c + p), except
possibly at c itself. Then lim f (x)< lim g(x).
x→c x→c
(b) Suppose that lim f (x) = 0 and lim [f (x)g(x)] = 1. Then
x→c x→c
lim g(x) dose not exist.
x→c
(a) That is not true bucause that let f (x) = 1 − x2 and
g(x) = 1 + x2, then lim f (x) = lim g(x).
x→c x→c
(b) Note that f (x) = 0 for all x ∈ (c−p, c+p), except possi-
bly at c itselfif because that f (x) = 0 and lim [f (x)g(x)] = 0.
x→c
It follows that lim g(x) is not exist because that lim [f (x)g(x)] =
x→c x→c
1
lim lim g(x) = 1.
x→c f (x) x→c
Calculus First Test 2011/10/20 5