Calculus I
Limits
Derivative
Exercices
Applications
Marginal analysis
© The scientific sentence. 2010

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 b^{f(x)} = b^{lim f(x)} = b^{c}
x → a
and if c > 0
lim log_{b} f(x) = log_{b} (lim f(x)) = log_{b} c
x → a
1.2. Examples
a)
lim 3^{2x  2} = 3^{ 2} = 1/9
x → 0
b)
lim log(4 x  1) = log 3
x → + 1
c)
lim e^{3x } = 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:
e^{lim 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 (e^{x})/(e^{x} + 1) = 1
x → + ∞
g)
lim f(x) = c
x → a
If c > 0,
Then
lim log_{b}f(x)
x → a
If c < 0,
Then
lim log_{b}f(x)
x → a
If c = 0,
Then
lim log_{b}f(x) = log_{b} lim f(x) = log_{b} 0^{+}
x → a
With
log_{b} 0^{+} =  ∞ if b > 1, and
log_{b} 0^{+} = + ∞ if 0 > b < 1,
Recall
if 0 < b < 1 we can write b = 1/B with B > 1
so log_{B}(x) = log_{b}(x) log_{B}(b)
(changing base formula)
We have : log_{B}(b) = log_{B}(1/B) =  log_{B}(B) =  1
so
log_{b}(x) =  log_{1/b}(x)
(1/b) > 1, then:
lim log_{1/b} 0^{+} =  ∞
x → a
so
lim log_{b} 0^{+} = + ∞
x → a
For any b > 0
lim log_{b} 0
x → a
Because
lim log_{b} 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) = log_{b} g(x) is continuous on the interval I
if g(x) > 0.
b) The function f(x) = b^{g(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} = 1^{1/0+} = 1^{+ ∞} = undetermined
x → 0^{+}
b) Limit at left:
lim (1 + x)^{1/x} = 1^{1/0} = 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 (e^{x}  1)/x = ?
x → 0
lim (e^{x}  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) = e^{x} or
x = e^{x}  1 or
1 = (e^{x}  1)/x
Therefore:
lim (e^{x}  1)/x = 1
x → 0

