###### Calculus I

Exercices

Applications

Marginal analysis

# 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 1e-05 (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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