Limits Of Sine Functions
Introduction to limits of sine function :-
Sine abbreviated as "sin" is one of the basic trigonometric functions giving the relation between the sides and and angles in a right triangle.Trigonometric functions basically relates the sides and angles of a triangle.
This article helps to learn limits of a sine function.
Explanation on Limits of Sine Function
Definition of sin function :-
Consider the right triangle ABC with right angle at B
limits of sine function
then sin (theta) is defined as ratio of AB/AC.
Properties of sin function :-
i)Between 0- 180o the sine function is + ve and from 180o to 360o the value is -ve.
ii) The sine function as value zero at 0o, 180o,360o.
Limits of sine function :-
i) Sine function has maximum value of 1 at (2n+1)90o and minimum of -1 at (2n+1) 270o where n is apositive integer.
Few important limits of sine function :-
i)limx->(pi/2) sinx = 1 as (sin (pi/2) =1)
ii)limx->(pi)sinx =0 as (sin pi =0)
iii)limx->(0)sinx =0 as (sin 0 =0)
iv)limx->(0)(sinx /x) =1 ( using lospitals rule ie differentiating numerator and denominator by x and applying limits)
v)limx->(infinity)(sinx ) = undefined (as sine wave oscillates at period of 360o )
Examples on Limits of Sine Function
Ex 1:-
Find limx->(0)(sin6x /4x)
Solution :-
sin6x /4x =(sin6x /6x)(6x/4x)
=> sin6x /4x = (sin6x /6x) 6/4
=>sin6x /4x = (sin6x /6x) 3/2
=>limx->(0)(sin6x /4x)=limx->(0)(sin6x /6x)(3/2)
=>limx->(0)(sin6x /4x) = 1.5 limx->(0)(sin6x /6x)
=>limx->(0)(sin6x /4x) = 1.5 ( as limx->(0)(sin6x /6x) =1 )
Ex 2:-
Find limx->(0)( 1+ sinx)
solution :
limx->(0)( 1+ sinx) =limx->(0) 1+ limx->(0) sinx
=>limx->(0)( 1+ sinx) = 1 (as limx->(0)sin x=sin0=0).
The trigonometric functions sine and cosine have four important limit properties:
You can use these properties to evaluate many limit problems involving the six basic trigonometric functions.
Example 1: Evaluate .
Substituting 0 for x, you find that cos x approaches 1 and sin x 3 approaches 3; hence,
Example 2: Evaluate
Because cot x = cos x/sin x, you find The numerator approaches 1 and the denominator approaches 0 through positive values because we are approaching 0 in the first quadrant; hence, the function increases without bound and and the function has a vertical asymptote at x = 0.
Example 3: Evaluate
Multiplying the numerator and the denominator by 4 produces
Example 4: Evaluate .
Because sec x = 1/cos x, you find that
Review Topics
Limits
Intuitive Definition
Evaluating Limits
One-sided Limits
Infinite Limits
Limits at Infinity
Limits Involving Trigonometric Functions
Continuity
The Derivative
Applications of the Derivative
Integration
Applications of the Definite Integral
Related Topics:
Differential Equations
Precalculus
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Limits Involving Trigonometric Functions
The trigonometric functions sine and cosine have four important limit properties:
You can use these properties to evaluate many limit problems involving the six basic trigonometric functions.
Example 1: Evaluate .
Substituting 0 for x, you find that cos x approaches 1 and sin x 3 approaches 3; hence,
Example 2: Evaluate
Because cot x = cos x/sin x, you find The numerator approaches 1 and the denominator approaches 0 through positive values because we are approaching 0 in the first quadrant; hence, the function increases without bound and and the function has a vertical asymptote at x = 0.
Example 3: Evaluate
Multiplying the numerator and the denominator by 4 produces
Example 4: Evaluate .
Because sec x = 1/cos x, you find that
Compute lim
x0
sin 7x
x
.
Plugging in x = 0 gives
0
0
. I have to do some more work.
The limit formula looks like this:
lim
0
sin
= 1.
(Im using instead of x to avoid confusing the variable in the formula with the variable in the problem.)
The point is that the thing that is going to 0, the thing inside the sine, and the thing on the bottom must be
identical.
by: johnharmer
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