MATH 1A Worksheet 3
October 7, 2014
Name:
Problem 1. You may use a calculator for these steps. This is to get an intuitive understanding of exponents and logarithms.
1) Graph ex and ln(x). What is the domain/range of these functions?
2) Explain in words and convince yourself of the following facts:
For any x, y ∈ R, we have ex+y = ex ey and exy = (ex )y = (ey )x
3) Calculate c = ln(2). Now, take the number c and calculate ec . In words, can you
explain what the function ln(x) returns when you feed in the input 2?
4) Now, explain in words and convince yourself of the following facts:
(a) ln(4) = ln(2) + ln(2)
(b) ln(6) = ln(2) + ln(3)
(c) ln(2/3) = ln(2) − ln(3)
(d) ln(2x ) = x ln(2) for any x
5) Finally, convince yourself that you can change bases via:
logb a =
logc a
for any new base c and old base b
logc b
(6) Graph 2x and log2 (x). Compare the graphs of (6) and (1). How are they related? (in
terms of horizontal/vertical stretches)
Problem 2. Solve for x:
(a) ln(ln(x)) = 1
Solution: ln(ln(x)) = 1 ⇒ ln(x) = e ⇒ x = ee
(b) ln(x) + ln(x + 1) = 2
Solution: Using the product rule:
⇒ ln(x(x + 1)) = 2
1
⇒ x(x + 1) = e2
⇒ x2 + x − e 2 = 0
√
−1 ± 1 + 4e2
⇒x=
2
We see that the negative solution is discarded, so x =
(c) e2x − 3ex + 2 = 0
Solution: Let u = ex , then
u2 − 3u + 2 = 0 ⇒ (u − 1)(u − 2) = 0 ⇒ u = 1, 2
⇒ x = ln(1), ln(2) = 0, ln(2)
(d) ln(x2 ) + log2 (x2 ) = 3
Solution: By change of bases,
ln(x2 ) +
ln(x2 )
=3
ln(2)
⇒ ln(x2 )(1 +
1
)=3
ln(2)
⇒ ln(x2 )(
ln(2) + 1
)=3
ln(2)
⇒ ln(x2 )(
ln(2e)
)=3
ln(2)
⇒ ln(x2 ) =
3 ln(2)
ln(2e)
3 ln(2)
⇒ x2 = e ln(2e)
3 ln(2)
⇒ x = ±e ln(2e)
Note to be careful when using ln(x2 ) = 2 ln x .
(e) (log2 x)(logx x2 ) = logx3 x2
Solution: By change of bases,
2
√
−1+ 1+4e2
.
2
ln x ln x2
ln x2
=
ln 2 ln x
ln x3
1
1
⇒
=
ln 2
ln x3
⇒ ln 2 = ln x3
⇒ 2 = x3 ⇒ x = 21/3
Problem 3. Compute the following limits (via squeeze):
(a) lim
x→∞
cos(tan(1/x))
ln(ln(x))
Solution: Since cos x ∈ [−1, 1],
−1
cos(tan(1/x))
1
≤
≤
ln(ln(x))
ln(ln(x))
ln(ln(x))
−1
cos(tan(1/x))
1
≤ lim
≤ lim
x→∞ ln(ln(x))
x→∞
x→∞ ln(ln(x))
ln(ln(x))
lim
Since lim
x→∞
lim
x→∞
1
−1
= 0 = lim
, we use the Squeeze Theorem to conclude that
x→∞
ln(ln(x))
ln(ln(x))
cos(tan(1/x))
=0
ln(ln(x))
(b) lim x2 sin(1/x)
x→0
Solution: Since sin x ∈ [−1, 1],
−x2 ≤ x2 sin(1/x) ≤ x2
⇒ lim −x2 ≤ lim x2 sin(1/x) ≤ lim x2
x→0
x→0
x→0
Since lim −x2 = 0 = lim x2 , we use the Squeeze Theorem to conclude that
x→0
x→0
lim x2 sin(1/x) = 0
x→0
(2x + sin(x))2
x→∞
x2
(c) lim
Solution: Since sin x ∈ [−1, 1], for x > 0,
(2x − 1)2
(2x + sin(x))2
(2x + 1)2
≤
≤
x2
x2
x2
3
(2x − 1)2
(2x + sin(x))2
(2x + 1)2
≤
lim
≤
lim
x→∞
x→∞
x→∞
x2
x2
x2
lim
(2x − 1)2
(2x + 1)2
=
4
=
lim
, we use the Squeeze Theorem to conclude that
x→∞
x→∞
x2
x2
(2x + sin(x))2
=4
lim
x→∞
x2
Since lim
(d) lim
x→∞
ln(x + 1) − ln(x)
2 + sin(x)
Solution: Note that ln(x + 1) − ln(x) = ln( x+1
) = ln(1 + 1/x), so
x
lim [ln(x + 1) − ln(x)] = lim ln(1 + 1/x) = ln(1) = 0
x→∞
x→∞
So, since 2 + sin(x) ∈ [1, 3], we see that
ln(x + 1) − ln(x)
ln(x + 1) − ln(x)
ln(x + 1) − ln(x)
≤
≤
3
2 + sin(x)
1
ln(x + 1) − ln(x)
ln(x + 1) − ln(x)
= 0 = lim
, by the Squeeze Theorem,
x→∞
3
1
ln(x + 1) − ln(x)
lim
=0
x→∞
2 + sin(x)
Since lim
x→∞
4