Proving: limit (as theta approaches zero) of sin(theta)/(theta) = 1

In summary, the author used the identity cos(2x) = cos2(x) - sin2(x) = 1 - 2sin2(h) to simplify the equation in line 1 to line 2. This allowed for the limit to be evaluated as -lim as theta approaches zero of ((sin\Theta/\Theta)*sin\Theta) which ultimately results in 0.
  • #1
einsteinoid
42
0

Homework Statement


show that the limit as h approaches zero of ((cosh-1)/h) = 0.

Homework Equations



*The bold numbers preceding each line are simply line numbers, they are not apart of the equations.*

1 lim (as h approaches 0) of ((cosh - 1)/h)

2 = lim (as h approaches 0) of -((2sin2(h/2))/h)

3 = -lim as theta approaches zero of ((sin[tex]\Theta[/tex]/[tex]\Theta[/tex])*sin[tex]\Theta[/tex]

4 = -(1)(0)

5 = 0
I'm confused as to how the author of the text was able to get from line 1 to line 2. If someone could workout the steps between these two lines I would be appreciative. I'm drawing a blank here!

Thanks
 
Physics news on Phys.org
  • #2
The author used this identity: cos(2x) = cos2(x) - sin2(x) = 1 - 2sin2(h), and then replaced 2x by h.
 
  • #3
Ah, I see. Thank you.
 

1. What is a limit?

A limit is a mathematical concept that describes the behavior of a function as its input approaches a certain value. It is used to determine the value that a function will approach, or "limit" to, as its input gets closer and closer to a particular value.

2. Why is the limit of sin(theta)/(theta) as theta approaches zero equal to 1?

This limit is a fundamental result in calculus known as the Sine Limit Theorem. It can be proven using the properties of trigonometric functions and the concept of a limit. As theta approaches zero, the value of sin(theta) also approaches zero, resulting in the limit of 1.

3. Can this limit be proven using algebraic manipulation?

Yes, the limit can be proven using algebraic manipulation. By using the trigonometric identity sin(theta) = theta - (theta^3)/3! + (theta^5)/5! - ..., the limit can be rewritten as (1 - (theta^2)/3! + (theta^4)/5! - ...) / theta. As theta approaches zero, the terms containing higher powers of theta become smaller and can be ignored, resulting in a limit of 1.

4. Is this limit important in real-world applications?

Yes, this limit is important in various fields such as physics, engineering, and economics. It is used to calculate the rate of change of a function, which is crucial in understanding the behavior of systems in the real world.

5. Are there any other important limits in calculus?

Yes, there are many important limits in calculus, including the Sine Limit Theorem mentioned above. Some other notable limits include the Power Rule for Limits, the Product Rule for Limits, and the Quotient Rule for Limits.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
275
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
632
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
23
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
980
  • Calculus and Beyond Homework Help
Replies
15
Views
5K
  • Calculus and Beyond Homework Help
Replies
24
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Replies
2
Views
180
Back
Top