Limit and Continuity of Trigonometric Functions
Continuity of Sine and Cosine function
Here is the graph of Sinx and Cosx
-We consider angles in radians
-Insted of θ we will use x
f(x) = sin(x)
g(x) = cos(x)
It is evident that as h approaches 0, the coordinate of P approach the corresponding coordinate of B.
But by definition we know
sin(0) = 0 and cos(0) = 1
The values of the functions matche with those of the limits as x goes to 0 (Remind the definition of continuity we have).
lim_{x → 0} sin(x) = sin(0) = 0 lim_{x → 0} cos(x) = cos(0) = 1
Hence we have the following theorem
DEFINITION 2.7.1
A function f(x) is said to be continuous at a point c if the following conditions are sarisfied
-f(c) is defined
-lim_{x → c} f(x) exists
-lim_{x → c} f(x) = f(c)
THEOREM 2.8.1
The functions sin(x) and cos(x) are continuous
Proof
Let h = x - c. So x = h + c. Then x → c is equivalent to the requirement that h → 0
A function f(x) is continuous at c if the following conditions are true:
-f(c) is defined
-lim_{h → 0} f(h + c) exists
-lim_{h → 0} f(h + c) = f(c)
Assume
lim_{x → 0} sin(x) = 0 and lim_{x → 0} cos(x) = 1
The first two conditions of continuity definition are met. We have to show that
lim_{h → 0} sin(c + h) = sin(c)
Now
lim_{h → 0} sin(c + h) = lim_{h → 0} [sin(c)cos(h) + cos(c)sin(h)] = lim_{h → 0} sin(c)cos(h) + lim_{h → 0} cos(c)sin(h) = sin(c)lim_{h → 0} cos(h) + cos(c)lim_{h → 0} sin(h) = sin(c)(1) + cos(c)(0) = sin(c)
Continuity of Other Trigonometric Functions
tan(x) = sin(x)/cos(x)
tan(x) is continuous everywhere except at cos(x) = 0 which gives
x = ± φ/2, ± 3φ/2, ± 5φ/2, ... = ± kφ/2 (k = 1, 3, 5, ...)
Similarly, since
cot(x) = cos(x)/sin(x)
sec(x) = 1/cos(x)
cosec(x) = 1/sin(x)
They are all continuous on appropriate ontervals using the continuity of sin(x) and cos(x).
Obtaining Limits by Squeezing
We will use Squeeze Theorem for finding limits
lim_{x → 0} sin(x)/x = 1
lim_{x → 0} [1 - cos(x)]/x = 0 Consider the graph of
Here the problem is:
- As x goes to 0, both thr top and the botton functions go to 0.
- sin(x) goes to 0 means that the fraction as a whole goes to 0.
- x goes to 0 means that the function as a whole goes to +∞.
Here we cannot write these functions in some other form by using algebraic Manipulation to solve this problem. So here we will use some other method. One such method is to obtain by following theorem:
Squeezing theorem
Let f, g and h be fucntions satisfying g(x)≤f(x)≤h(x) for all x in some open interval containg the point a, whit the possible exception that the inequalities need not hold at a.
If g and h have the same limits as x approaches a, say
lim_{x → a} g(x) = lim_{x → a} h(x) = L
then f also has this limit as x approaches a, that is
lim_{x → a} f(x) = L
Example:
Use the squeezing theorem to evaluate
lim_{x → 0} x^{2}sin^{2}(1/x)
Solution
Remember that 0 ≤ sin(x) ≤ 1, so 0 ≤ sin^{2}(x) ≤ 1 and so 0 ≤ sin^{2}(1/x) ≤ 1
Multiply throughtout this last inequality by x^{2}
0 ≤ x^{2}sin^{2}(1/x) ≤ x^{2}
But lim_{x → 0} 0 = lim_{x → 0} x^{2} = 0
So by the Squeezing Theorem
lim_{x → 0} x^{2}sin^{2}(1/x) = 0
Before proving next theorem, let us see the following formula.
The proof will use basic facts about circles and areas of sectors with angle θ radians and raius r
The area of a sector is given by
A = (1/2).r^{2}θ
THEOREM 2.8.3
lim_{x → 0} sin(x)/x = 1
Let x be such that 0 < x < φ/2 or -φ/2 < x < 0
We made the assumption that
0 < x < φ/2
Also works when -φ/2 < x < 0
From the figure
0 < area of ΔOBP = (1/2) base.height = (1/2) (1).sin(x) = (1/2) sin(x)
area of sector OBP = (1/2)(1)^{2}.x = (1/2)x
area of ΔOBQ = (1/2) base.height = (1/2) (1).tan(x) = (1/2)tan(x)
So, the above inequality becomes
0 < (1/2)sin(x) < (1/2)x < (1/2)tan(x)
Multiplying throughout by 2/sin(x)
1 < x/sin(x) < 1/cos(x)
Taking resiprocal gives
cos(x) < sin(x)/x < 1
Taking limit now and using Squeezing Theorem gives
lim_{x → 0} cos(x) < lim_{x → 0} sin(x)/x < lim_{x → 0} 1 = 1 < lim_{x → 0} sin(x)/x < 1
Since the middles term is between 1 and 1, it must be 1