Calculus I
Limits
Derivative
Exercices
Applications
Marginal analysis
© The scientific sentence. 2010
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Calculus I: exponential function
logarithmic function and continuity
1. Limits: exponential &
logarithmic functions
1.1. Definitions
All the rules of limits that apply to algebraic functions apply
also to the exponential functions and logarithmic functions.
In addition, we have:
With a and c real numbers or infinite (± ∞),
b positive real number and not equal to 1,
If
lim f(x) = c
x → a
Then
lim bf(x) = blim f(x) = bc
x → a
and if c > 0
lim logb f(x) = logb (lim f(x)) = logb c
x → a
1.2. Examples
a)
lim 32x - 2 = 3- 2 = 1/9
x → 0
b)
lim log(4 x - 1) = log 3
x → + 1
c)
lim e3x = e- ∞ = 0
x → - ∞
d)
lim (1/3)x = (1/3)+ ∞ = 1+ ∞/3+ ∞
x → + ∞
= (undetermined) /∞
Let's write:
y = (1/3)x
so
ln y = x ln (1/3) = - x ln 3
Then:
lim ln y = lim (- x ln 3) = - ∞
x → + ∞
Therefore:
elim ln y = lim e ln y = lim y = e- ∞ = 0
x → + ∞
Hence
lim (1/3)x = 0
x → + ∞
e)
lim x (1 - log (4x)) = ∞ (- ∞) = - ∞
x → + ∞
f)
lim (ex)/(ex + 1) = 1
x → + ∞
g)
lim f(x) = c
x → a
If c > 0,
Then
lim logbf(x) 
x → a
If c < 0,
Then
lim logbf(x) 
x → a
If c = 0,
Then
lim logbf(x) = logb lim f(x) = logb 0+
x → a
With
logb 0+ = - ∞ if b > 1, and
logb 0+ = + ∞ if 0 > b < 1,
Recall
if 0 < b < 1 we can write b = 1/B with B > 1
so logB(x) = logb(x) logB(b)
(changing base formula)
We have : logB(b) = logB(1/B) = - logB(B) = - 1
so
logb(x) = - log1/b(x)
(1/b) > 1, then:
lim log1/b 0+ = - ∞
x → a
so
lim logb 0+ = + ∞
x → a
For any b > 0
lim logb 0
x → a
Because
lim logb 0-
x → a
2. Continuity: exponential &
logarithmic functions
2.1. Definitions
If the function g(x) is continuous on an interval I, then
a) The fonction f(x) = logb g(x) is continuous on the interval I
if g(x) > 0.
b) The function f(x) = bg(x) is continuous on the interval I.
2.2. Examples
a)
The polynomial function g(x) = 3x - 3 is continouous on R.
It is positive when x > 1, so
f(x) = ln (3x - 3) is continuous on the interval R with
the condition x > 1, that is on the interval ]1, + ∞[
b)
f(x) = e(3x - 3) is continuous on the interval R.
c)
f(x) = log ((x + 1)/(x - 1))
is continuous on the interval where (x + 1)/(x - 1) is
positive, that is on I = ]- ∞, -1[ U ]+1, + ∞[
3. Exercises
1)
What is
lim (1 + x)1/x = ?
x → 0
lim (1 + x)1/x = 1/0
x → 0
a) Limit at right:
lim (1 + x)1/x = 11/0+ = 1+ ∞ = undetermined
x → 0+
b) Limit at left:
lim (1 + x)1/x = 11/0- = 1- ∞ = undetermined
x → 0-
Now fill in the two following tables:
| x | 1 | 0.1 | 0.01 | 0.001 | 0.0001 | 0.00001 |
| (1 + x)1/x | 2 | 2.5937 | 2.7048 | 2.7169 | 2.7181 | 2.7182 |
| x | - 1 | - 0.1 | - 0.01 | - 0.001 | - 0.0001 | - 0.00001 |
| (1 + x)1/x | 0 | 2.8686 | 2.7322 | 2.7196 | 2.7184 | 2.71829 |
The more x approaches 0, the more f(x) approaches
the value of the Euler number e = 2,71828.
Therefore:
lim (1 + x)1/x = e
x → 0
2)
lim (ex - 1)/x = ?
x → 0
lim (ex - 1)/x = 0/0 = undetermined
x → 0
We know
lim (1 + x)1/x = e
x → 0
so, at the neighborhood of 0:
(1 + x)1/x = e or
(1 + x) = ex or
x = ex - 1 or
1 = (ex - 1)/x
Therefore:
lim (ex - 1)/x = 1
x → 0
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