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Test # 1, September 11 th, 2007, Solutions Math 1131
- Determine the following limits numerically and analytically:
(a) lim x→∞
2 x + 1 2 x − 3
) 3 x (b) lim x→ 0
1 − cos 5x x^2
(c) lim x→ 1
√ (^5) x − 1 √ (^3) x − 1
Solution: (a) Let f (x) =
( 2 x+ 2 x− 3
) 3 x for x > 2. Some of the values of f at inputs that are getting larger and larger are tabulated next: f(10) 566. f(100) 415. f(1000) 404. f(10000) 403. f(100000) 403. f(1000000) 403. It seems to be the case that the limit is ≈ 403 .4. To do this algebraically we use the first fundamental limit:
lim x→∞
2 x + 1 2 x − 3
) 3 x = lim x→∞
2 x − 3
) 3 x = lim x→∞
[(
(2x − 3)/ 4
) 2 x 4 − 3 ](3 2 xx−)4 3 = e
lim x→∞
2 − 3 /x (^) = e^6
so
lim x→∞
2 x + 1 2 x − 3
) 3 x = e^6 ≈ 403. 42879349 ....
(b) If g(x) =
1 − cos 5x x^2
if x 6 = 0. Some of the values of g for inputs getting closer and
closer to zero are included in the next table: g(-0.1) 12. g(0.01) 12. g(-0.001) 12. g(0.0001) 12. g(-0.00001) 12. g(0.000001) 12. We can guess that this limit must be equal to 12.5 or 252. This is indeed the cases since
lim x→ 0
1 − cos 5x x^2
= lim x→ 0
2 sin^2 (^52 x ) x^2
= 2 lim x→ 0
sin 52 x 5 x 2
Hence lim x→ 0
1 − cos 5x x^2
I
(c) Finally if h(x) =
√ (^5) x − 1 √ (^3) x − 1 for every real number x. Some of the values of g around
zero are shown below: h(0.9) 0. h(1.01) 0. h(0.999) 0. h(1.0001) 0. h(0.99999) 0. h(1.000001) 0. So it is reasonable to conclude that the limit of this function at x = 1 is 0.6 = 3/5. This is true since if we make the change of variable x = (1 + t)^3 we see that t → 0 and
lim x→ 1
√ (^5) x − 1 √ (^3) x − 1 = lim t→ 0
(1 + t) (^35) − 1 t
Therefore, lim x→ 1
√ (^5) x − 1 √ (^3) x − 1 = 3/ 5. •
- Determine if the following functions are continuous or not. If they are not continuous find the points of discontinuity.
(a) f (x) =
x^2 + x if x ≥ 0 sin 3x x
if x < 0
, (b) g(x) =
ln(^1 e + x) if x ≥ 0 ln(1 + |x|) x
if x < 0
Solution: (a) The function f is continuous at every point except maybe x = 0. At x = 0
we have lim x→ 0 +^
x^2 + x = 0 = f (0). and lim x→ 0 −^
f (x) = lim x→ 0 −
sin 3x x
= 3 lim x→ 0 −
sin 3x 3 x
= 3. Therefore
the function f is discontinuous at x = 0 but continuous at every other point. (b) The function g is continuous at every point except maybe x = 0. The limit from the right hand side of 0 is lim x→ 0 +^
ln(1/e + x) = ln 1/e = −1 = g(0). For the limit from the left at
0 we get
lim x→ 0 −^
g(x) = lim x→ 0 −
ln(1 + |x|) x
= lim x→ 0 −
ln(1 + |x|) |x|
|x| x
So g is continuous. •
- Find all values of a such that the following function is continuous:
h(x) =
ax 2 + a^2 x
if x ≥ 1
√ (^3) x − 1
x − 1
if x < 1
II
which gives h′(x) =
2 x + 3
, x 6 = −
(b) Similarly
i′(a) = lim x→a
i(x) − i(a) x − a
= lim x→a
x+ 2 x− 3 −^
a+ 2 a− 3 x − a
= lim x→a
(x + 1)(2a − 3) − (a + 1)(2x − 3) (x − a)(2x − 3)(2a − 3)
lim x→a
(2x − 3)(2a − 3)
(2a − 3)^2
which gives i′(x) = −
(2x − 3)^2
(c, Bonus:) We have
k′(a) = lim x→a
k(x) − k(a) x − a
= lim x→a
x^2 x− 1 −^
a^2 a− 1 x − a
= lim x→a
x^2 (a − 1) − a^2 (x − 1) (x − a)(x − 1)(a − 1)
lim x→a
ax(x − a) − (x − a)(x + a) (x − a)(x − 1)(a − 1)
= lim x→a
ax − x − a (x − 1)(a − 1)
a^2 − 2 a (a − 1)^2
and so k′(x) =
x(x − 2) (x − 1)^2
, x 6 = 1..
- For the functions in Problem 5 find the equation of the tangent line at x = 3 and plot both the tangent line and the corresponding function.
Solution: (a) The equation of the tangent line is y = h(3) + h′(3)(x − 3). Since h(3) = 3
and h′(1) = 1/3 we get y = 3 + (x − 3)/3 or y = (x + 6)/3. The graph of h and of the
tangent line is included below:
4
3
x
2
1
(^00246)
(b) Since i(3) = 4/3 and i′(3) = −^59 the equation of the tangent line at (3, 4 /3) is
y =
27 − 5 x 9
. The graph is
IV
x
2
3
4
1
2.5 3.5 5
2 3
(c) In this case k(3) = 9/2 and k′(3) = 3/4. Then the equation of tangent line becomes
y = 9/2 + 3(x − 4)/4 or y = (3x + 9)/ 4 and the graph:
4.5 5
6
3
x
4
5
2 2.
4
- Use the Intermediate Value Theorem to show that the following equation has a solution in the specified interval:
tan 2x = x in (
π 4
3 π 4
Solution: Consider the function f (x) = sin 2x−x cos 2x defined for all real x. All elementary functions involved are defined for all real numbers. Hence f is continuous on [π 4 , 34 π ]. Since f (π 4 ) = sin π 2 − π 4 cos π 2 = 1 and f (^34 π ) = sin 32 π − 34 π cos 32 π = −1 we see that 0 is between f (π 4 ) and f (^34 π ). Hence by IVT we know there is a number x ∈ (π 4 , 34 π ) such that f (x) = 0. This means sin 2x − x cos 2x = 0 or tan 2x = x. •
- A particle moves along a straight line with equation of motion s(t) = 4t^3 − 15 t^2 + 12 t + 7, where s is measured in meters and t in seconds. Find the time when the velocity becomes zero.
Solution: We calculate the speed v := dsdt = 12t^2 − 30 t+12 = 6(2t^2 − 5 t+2) = 6(2t−1)(t−2). Therefore, v(t) = 0 implies t = 0.5 seconds and t = 2 seconds. •
- Let f (x) = | sin(x)| defined for all real numbers x. Where is f differentiable? What about g(x) = | sin(x)|^3?
V
