Math 180: Calculus I
Fall 2014
September 23
TA: Brian Powers
1. Evaluate the following limits using the facts that:
lim
x→0
sin x
=1
x
and
lim
x→0
cos x − 1
=0
x
(a) limx→0 tanx5x
SOLUTION:
lim
x→0
tan 5x
sin 5x
= lim
definition of tan
x→0 x cos x
x
1
sin 5x
= lim
lim
Limit product rule
x→0 cos x
x→0
x
sin 5x
= (1) lim
Evaluate first limit
x→0
x
sin 5x
5
= 5 lim
multiply by
x→0 5x
5
sin u
= 5 lim
use substitution u = 5x
u→0 u
= 5(1) = 5
(b) limx→−3 xsin(x+3)
2 +8x+15
SOLUTION:
lim
x→−3
sin(x + 3)
sin(x + 3)
= lim
factor denominator
x2 + 8x + 15 x→−3 (x + 3)(x + 5)
sin(u)
= lim
use substitution u = x + 3
u→0 (u)(u + 2)
sin u
1
= lim
lim
limit product rule
u→0 u
u→0 u + 2
1
1
= (1)
=
2
2
2. Evaluate the derivatives dy/dx
cos x
(a) y = sin
x+1
SOLUTION: Use the quotient rule with u = cos x, v=sin x + 1, so u0 = − sin x, v 0 = cos x.
dy
u0 v − v 0 u
=
dx
v2
(− sin x)(sin x + 1) − (cos x)(cos x)
=
(sin x + 1)2
7-1
7-2
TA Discussion 7: September 23
(b) y = csc x using the quotient rule
SOLUTION: First rewrite csc x =
1
sin x ,
so u = 1, v = sin x, u0 = 0, v 0 = cos x0 .
dy
u0 v − v 0 u
=
dx
v2
(0)(sin x) − (cos x)(1)
=
sin2 x
cos x
=− 2
sin x
cos x 1
=−
sin x sin x
= − cot x csc x
cot x
(c) y = 1+csc
x
SOLUTION: Quotient rule: u = cot x, v = 1 + csc x, u0 = − csc2 x, v 0 = − cot x csc x.
dy
u0 v − v 0 u
=
dx
v2
(− csc2 x)(1 + csc x) − (− cot x csc x)(cot x)
=
(1 + csc x)2
cos x
(d) y = x1+x
3
SOLUTION: Quotient rule with u = x cos x, u0 = −x sin x + cos x (by product rule),
v = 1 + x3 , v 0 = 3x2 .
dy
u0 v − v 0 u
=
dx
v2
(−x sin x + cos x)(1 + x3 ) − (3x2 )(x cos x)
=
(1 + x3 )2
3. Evaluate the following limit or state it does not exist.
lim
x→0
sin ax
, where a and b are constants with b 6= 0
sin bx
SOLUTION:
sin ax
sin ax abx
lim
= lim
multiply by 1
x→0 sin bx
x→0 sin bx
abx
a sin ax
bx
= lim
lim
limit product rule
x→0
x→0 b sin bx
ax
sin u
1
v
= a lim
lim
let u = ax, v = bx
u→0 u
b v→0 b sin v
−1
a
b sin v
= (1) lim
= evaluate limit and use limit exponent rule
v→0
b
v
a
= (1)−1
evaluate limit
b
4. Find the following derivatives using the product rule
d
(sin2 x)
dx
d
(sin3 x)
dx
d
(sin4 x)
dx
TA Discussion 7: September 23
Make a conjecture about
SOLUTION:
7-3
n
d
dx (sin
x). See if you can prove it by induction!
d
d
(sin2 x) =
(sin x sin x) = (sin x cos x) + (cos x sin x) = 2 cos x sin x
dx
dx
d
d
(sin3 x) =
(sin x sin2 x) = (sin x(2 cos x sin x)) + (cos x sin2 x) = 3 cos x sin2 x
dx
dx
d
d
(sin4 x) =
(sin x sin3 x) = (sin x(3 cos x sin2 x)) + (cos x sin3 x) = 4 cos x sin3 x
dx
dx
We may conjecture that
d
(sinn x) = n cos x sinn−1 x
dx
Proof by induction:
d
(sinn x) = n cos x sinn−1 x.
We have already shown the base case for n = 2, 3, 4. Assume for n ≥ 2 that dx
n+1
(n+1)−1
d
We wish to show that dx
(sin
x) = (n + 1) cos x sin
.
d
d
(sinn+1 x) =
(sin x sinn x)
dx
dx
factor out a sin x
d
(sinn x))
product rule
dx
n
n−1
= cos x sin x + sin x(n cos x sin
x)invoke inductive hypothesis
= (cos x sinn x) + (sin x
= cos x sinn x + n cos x sinn x
= (n + 1) cos x sin(n+1)−1 x
Done! It’s worth noting that this is simply a special case of the Chain Rule, a very powerful rule for
derivatives.
d
5. Use the fact that cos(x + h) = cos(x) cos(h) − sin(x) sin(h) to prove that dx
cos x = − sin x using the
limit definition.
SOLUTION:
d
cos(x + h) − cos x
cos x = lim
h→0
dx
h
cos(x) cos(h) − sin(x) sin(h) − cos x
= lim
h→0
h
cos x cos h − cos x
sin x sin h
= lim
− lim
h→0
h→0
h
h
sin h
cos h − 1
= cos x lim
− sin x lim
h→0
h→0 h
h
= cos x(0) − sin x(1)
= − sin x