###### Calculus I

Exercices

Applications

Marginal analysis

# Calculus I: Trigonometric functions Limits & continuity

### 1. Limit of Trigonometric functions

##### 1.1. Definitions

1.
If lim f(x) = b (a is real or infinite, and b real)
x → a

Then
a)
 lim sin f(x) = sin [lim f(x)] = sin b x → a x → a

b)
 lim cos f(x) = cos [lim f(x)] = cos b x → a x → a

2.
If lim f(x) = ± ∞ (a is real or infinite)
x → a

Then

a)
 lim sin f(x) x → a

b)
 lim cos f(x) x → a

##### 1.2. Two particular cases:

lim (sin x)/x = 1
x → 0

Proof: see Exercises section.

lim (cos x - 1)/x = 0
x → 0

Proof: see Examples section.

##### 1.3. Examples

a)
lim sin (3x) = 0
x → 0

b)
lim cos (3x/(x + 1)) = cos 3
x → + ∞

c)
lim tg (2x) = tg (π) = 0
x → + π/2

d)
lim cos (2x)
x → + ∞

e)
lim sin(x)/x = ?
x → + ∞

lim sin(x)/x = /∞ : ambiguous!
x → + ∞

Remark that
lim sin(x) = real number in the interval [- 1 , + 1]
x → + ∞

Since
lim 1/x = 0
x → + ∞

We find:

lim sin(x)/x = 0 x real number in the interval [- 1 , + 1] = 0
x → + ∞

lim sin(x)/x = 0
x → + ∞

lim sin(x)/x = 0
x → + ∞

f) f(x) = sin3(x)/5x3

lim sin3(x)/5x3
x → + ∞
= (1/5) lim [sin(x)/x]3 = (1/5) (1)3 = 1/5
x → + ∞

g)
f(x) = ((cos x) - 1)/x

lim ((cos x) - 1)/x = 0/0: undetermined
x → 0

We have:

((cos x) - 1)/x = ((cos x) + 1)((cos x) - 1)/x ((cos x) + 1) =
((cos2 x) - 1)/x ((cos x) + 1) = (- sin2 x)/x ((cos x) + 1) =
- (sin x)/x . (sin x)/((cos x) + 1)

Therefore:

lim ((cos x) - 1)/x
x → 0

=
lim [- (sin x)/x . (sin x)/((cos x) + 1)] = (- 1). (0/2) = 0
x → 0

lim ((cos x) - 1)/x = 0
x → 0

### 2. Continuity of trigonometric functions

##### 2.1. Definitions

The two functions sin(x) and cos(x) are continuous on R = ]- ∞, + ∞[.

If a function g(x) is continuous on an interval I, then the function f(x) = sin g(x), and f(x) = cos g(x) are continuous on the interval I.

##### 2.2. Examples

Define the continuity of the following functions: a)
f(x) = cos (x - 1)1/2

The domain of the function g(x) = (x - 1)1/2 is I = ]1, + ∞[. So g(x) is continuous on the interval I, so does the function f(x).

b)
f(x) = tan(x) = sin x/cos x
f(x) is discontinuous for the values of x such as cos x = 0:
x = ± (2 k + 1)π/2
k is an integer.

### 3. Exercises

##### 3.1. Exercise 1: (sin x)/x

f(x) = (sin x)/x

lim (sin x)/x = 0/0 : Undetermined
x → 0

Use a calculator in MODE radian, and fill in the following two tables to find:

lim (sin x)/x = 1
x → 0+

and

lim (sin x)/x = 1
x → 0-

1. x approaches 0 from the left:

 x → 0- f(x) - 1 0.8414 - 0.5 0.9588 - 0.1 0.9983 - 0.05 0.9995 - 0.001 0.99999

2. x approaches 0 from the right:

 0+ → x f(x) 1 0.8414 0.5 0.9588 0.1 0.9983 0.05 0.9995 0.001 0.99999

Conclude that:

lim (sin x)/x = 1
x → 0

##### 3.2. Exercise 2: sin(3x)/tan(4x)

f(x) = sin(3x)/tan(4x)

We want to evaluate:

lim f(x) = sin(3x)/tan(4x)
x → 0

The value of 0 is: f(0) = 0/0 which is undetermined.

We have:
sin(3x) = sin(4x - x) = sin(4x) cos(x) - cos(4x) sin(x)

We know also: tan(4x) = sin(4x)/cos(4x)

We obtain f(x) =

(sin(4x) cos(x) - cos(4x) sin(x))/(sin(4x)/cos(4x)) =
cos(4x)) x [cos(x) - (cos(4x) sin(x))/sin(4x)]

We know also:

sin(4x) = 2 sin(2x) cos(2x) = 4 sin(x) cos(x) cos (2x).

So

f(x) = cos(4x) [cos(x) - cos(4x)/4 cos(x) cos(2x)]

Then

lim f(x) = lim {cos(4x) [cos(x) - cos(4x)/4 cos(x) cos(2x)]} = 1 [1 - 1/4] = 3/4.
x → 0

lim sin(3x)/tan(4x) = 3/4.
x → 0

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