HW II ,Answers
1)a)Find an open interval about x 0 = 5 on which the inequality
9 − x − 2 < ε holds
for any ε > 0 .
9 − x − 2 < ε ⇔ −ε < 9 − x − 2 < ε
⇔ 2−ε < 9−x < ε+2
⇔ (2 − ε) 2 < 9 − x < (ε + 2) 2 if 0 < ε ≤ 2.
⇔ 5 − 4ε − ε 2 < x < 5 + 4ε − ε 2
9 − x − 2 < ε whenever x ∈ I = (5 − 4ε − ε 2 ,5 + 4ε − ε 2 ), x ≠ 5.
b)What is the largest possible value for δ > 0 such that
9 − x − 2 < ε whenever
0 < x −5 < δ.
max δ = min {4ε − ε 2 , 4ε + ε 2 } = 4ε − ε 2 .
ε > 2 ⇒ 0 < 9 − x < (2 + ε) 2
⇒ x ∈ (5 − 4ε − ε 2 ,9)
Thus, δ ≤ 4.
c)Did you complete the proof of lim 9 − x = 2 ? Yes.
x →5
2)Find the following limits if they exist
a) lim
s→2
2
3
3
(2 − s)(4 + 3s + 10)
= lim
s →2
b) lim
x →∞
2
= .
9
(
2
3
1
3
3(2 − s)((10 − s) + 2(10 − s) + 4)
)
x 2 + 3x + 3 − x = lim( x 1 + 3x −1 + 3x −2 − x)
x →∞
= lim x( 1 + 3x −1 + 3x −2 − 1)
x →∞
= ∞.0 undetermined , but
(
lim ( x + 3x + 3 − x ) = lim ( x + 3x + 3 − x ) .
(
2
2
x →∞
1
3
10 − s − 2
10 − s − 2 4 + 3s + 10 (10 − s) + 2(10 − s) + 4
= lim
2
1
s
2
→
4 − 3s + 10
4 − 3s + 10 4 + 3s + 10
3
(10 − s) + 2(10 − s) 3 + 4
3
x →∞
= lim
x →∞
= lim
x →∞
(
( 3x + 3)
x 2 + 3x + 3 + x
)
3x + 3
x 1 + 3x −1 + 3x −2 + x
)
+ 3x + 3 + x )
x 2 + 3x + 3 + x
x2
= lim
x →∞
=
c) lim
x →−∞
(
x(3 + 3x −1 )
x( 1 + 3x −1 + 3x −2 + 1)
3
.
2
)
x 2 + 3x + 3 − x = = lim( x 1 + 3x −1 + 3x −2 − x)
x →∞
= lim(− x)( 1 + 3x −1 + 3x −2 + 1)
x →∞
= ∞ ,i.e.,limit does not exist.
sin 8x −
sin 8x
cos8x
sin 8x − tan 8x
= lim
x →0
x3
x3
sin 8x(cos 8x − 1)
= lim
x →0
x 3 cos 8x
(cos 8x − 1)
8sin 8x
1
= lim
lim
lim
2
x →0
x →0
x → 0 cos 8x
x
8x
= ( −32)(8)(1) = −256, where
(cos8x − 1) (cos8x + 1)
(cos8x − 1)
= lim
lim
2
x →0
x →0
x2
(cos8x + 1)
x
d) lim
x →0
sin 2 8x
1
= − lim
2
x →0
x
cos8x + 1
sin 8x 2
1
= −(lim
) .64.
x →0
8x
cos8x + 1
= −32.
⎛ x3
x2 ⎞
2x 3 + 4x 2
2
−
= .
e) lim ⎜ 2
=
lim
⎟
2
x →∞ 3x − 4
3x + 2 ⎠ x →∞ ( 3x − 4 ) ( 3x + 2 ) 9
⎝
3)Let
if x ≤ −1
⎧ mx + n
⎪
g(x) = ⎨
⎪mx 3 + x + 2n if x > −1.
⎩
a)Find the values of m and n so that g is differentiable at x=-1.
Continuity of g( x) :
lim g(x) = lim− (mx + n) = − m + n, and lim+ g(x) = lim+ (mx 3 + x + 2n) = −m − 1 + 2n, and
x →−1−
x →−1
x →−1
lim g(x) = lim+ g(x) = g(−1) ⇒ n = 1.So,
x →−1−
x →−1
, x ≤ −1
⎧ mx + 1
g(x) = ⎨ 3
⎩mx + x + 2 , x > −1
is continuous at x = −1.
g( −1 + h) − g(−1)
g′( −1) = lim
h →0
h
x →−1
where
g( −1 + h) − g(−1)
m(h − 1)3 + (h − 1) + 2 − (1 − m)
= lim+
= 3m + 1,
h →0
h →0
h
h
g( −1 + h) − g(−1)
m(h − 1) + 1 − (1 − m)
g′− (−1) lim−
= lim−
= m.
h →0
h →0
h
h
−1
−1
g′+ (−1) = g′− (−1) ⇔ m = , and g′(−1) = .
2
2
−1
Thus, g is differentiable at x = −1 if m = , n = 1.
2
b)For the values you found in part a),find g′(x).
g′+ (−1) = lim+
⎧ −1
if x ≤ −1
⎪⎪
2
g′(x) = ⎨
⎪ −3 x 2 + 1 if x > −1.
⎪⎩ 2
c)Is g′(x) differentiable at x=-1.
g′(x) is continuous at x=-1 but
d
g′( −1 + h) − g′(−1)
does not
g′(x) x =−1 = lim
h →0
dx
h
g′(−1 + h) − g′(−1)
= −3
h →0
h
WHY?).Thus, g′(x) is not differentiable at x=-1.
exist(i.e.,
g′(−1 + h) − g′(−1)
=0
lim
h → 0−
h
lim+
4)a)Using the definition of the derivative find
dy
if y = 6 cos 7x.
dx
d
6 cos(7(x + h)) − 6 cos 7x
(6 cos 7x) = lim
h →0
dx
h
6(cos 7x cos 7h − sin 7x sin 7h − cos 7x)
= lim
h →0
h
cos 7h − 1
7 sin 7h
= (6 cos 7x)(lim
) − (6sin 7x)(lim
)
h →0
h →0
h
7h
cos 7h − 1
sin 7h
= −42 sin 7x, where lim
= 0, and lim
= 1.
h →0
h →0
h
7h
f (g(x)) − f (2)
b)Evaluate lim
where f and g are differentiable functions with
x →−1
x +1
f ′(2) = 3, g( −1) = 2, g′( −1) = −5.
f (g(x)) − f (g(−1)) g(x) − g(−1)
f (g(x)) − f (2)
= lim
lim
x →−1
x →−1
x +1
g(x) − g(−1)
x +1
f (g(x)) − f (g(−1))
g(x) − g(−1)
. lim
= lim
x →−1
x
→−
1
g(x) − g(−1)
x +1
f (z) − f (2))
g(x) − g(−1)
= lim
,where z = g(x)
. lim
z→2
x
→−
1
z−2
x +1
= f ′(2)g′( −1) = −15.
⎛ π − π cos x ⎞
x sin ⎜
⎟
x
⎝
⎠ is not continuous at x=0.Is the discontinuity
5)The function f (x) =
1 − cos x
removable?If so,what would f(0) be?
⎡ ⎛ 1 − cos x ⎞ ⎤
sin ⎢ π ⎜
⎟⎥
x
π sin πu
1 − cos x
⎝
⎠⎦
⎣
L = lim
= lim
= π, where u =
, and u → 0 as
x →0
u →0
1 − cos x
πu
x
x
x → 0. Thus ,the discontinuity is removable and f(0)= π.
6)An automobile travels without refueling until it runs out of gasoline.Prove that at
some point on the journey,the number of liters of fuel tank is equal to the number of
kilometers traveled.
Let k(t) be the number of kilometers traveled,and let l(t) be the number of liters of
fuel in the tank.Since k starts at 0 and ends positive,and l starts positive and ends
at 0,the function d (t) = k(t) − l(t) < 0 at the beginning and d(t) = k(t) − l(t) > 0 at the
end.Therefore,at some point t 0 , d(t 0 ) = 0 ,and k(t 0 ) = l(t 0 ) (nothing but the
Intermediate Value Theorem)
7)How many lines are tangent to both of the parabolas y = 1 + x 2 and y = −1 − x 2 .Find
the coordinates of the points at which these tangents touch the parabolas.
Let P be a point at which one of these tangents touches the upper parabola and let
a be its x-coordinate.Then,since P lies on the parabola y = 1 + x 2 ,its y-coordinate
must be 1 + a 2 .Because of the symmetry,the coordinates of the point Q where the
1+ a2
. If
tangent touches the lower parabola must be Q (−a 2 , −(1 + a 2 )). We have m PQ =
a
f (x) = 1 + x 2 ,then the slope of the tangent line at P (a 2 ,1 + a 2 ) is
f (a + h) − f (a)
(1 + (a + h) 2 ) − (1 + a 2 )
= lim
= 2a. Thus,the condition that we
h →0
h →0
h
h
1+ a2
need to use is that 2a =
which implies a = 1.∴The points are (1, 2) and
a
(−1, −2). By symmetry,the two remaining points are (−1, 2) and (1, −2).
Or(without using symmetry) Let P(a,1 + a 2 ) be a point at which one of these
tangents touches the upper parabola y = 1 + x 2 and Q(b, −(1 + b 2 )) be a point at which
one of these tangents touches the lower parabola y = −(1 + x 2 ) .Then,
dy
y = 1+ x2 ⇒
= 2a,
dx P
f ′(a) = lim
y − −(1 + x 2 ) ⇒
⇒ b = −a.
dy
= −2b
dx Q
Thus, m PQ =
1 + a 2 − ( −1 − b 2 ) a 2 + 1
=
,and the condition that we need to use is that
a−b
a
1+ a2
2a =
which implies a = 1.∴ The points are (1, 2) , (−1, −2) , (−1, 2) and (1, −2).
a