This document contains the work of a student on a calculus test. It includes:
1) Solving limits, finding derivatives, and applying L'Hopital's rule.
2) Using induction to prove an identity.
3) Providing epsilon-delta proofs of limits.
4) Finding where a tangent line is parallel to a secant line.
5) Proving statements about limits of functions.
The student provides detailed solutions showing their work for each problem on the test.
Using implicit differentiation we can treat relations which are not quite functions like they were functions. In particular, we can find the slopes of lines tangent to curves which are not graphs of functions.
Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.
Using implicit differentiation we can treat relations which are not quite functions like they were functions. In particular, we can find the slopes of lines tangent to curves which are not graphs of functions.
Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.
This is meant for university students taking either information technology or engineering courses, this course of differentiation, Integration and limits helps you to develop your problem solving skills and other benefits that come along with it.
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