Evaluating a limit to 2 decimal places

  • Thread starter meeklobraca
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In summary, the problem is asking to evaluate the limit of (f(2+h) - f(2))/h as h approaches 0, where f(x) = sin x. After trying different values for h, it is found that the limit is equal to cos 2, or approximately -0.42. The method used to solve this problem depends on the context and can vary.
  • #1
meeklobraca
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Homework Statement



If f(x) = sin x, evaluate

lim f(2+h) - f(2) / h
h->0

Evaluate to 2 decimal places


Homework Equations





The Attempt at a Solution



I think that since f(2) then the answer is sin 2 which is .91

What do you guys think?
 
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  • #2
meeklobraca said:

Homework Statement



If f(x) = sin x, evaluate

lim f(2+h) - f(2) / h
h->0

Evaluate to 2 decimal places

Homework Equations


The Attempt at a Solution



I think that since f(2) then the answer is sin 2 which is .91

What do you guys think?

If it said lim sin(h) as h->2, then the answer would be sin 2 since sin(x) is continuous.

However, the limit expression is not sin(x), it is f(2+h) - f(2) / h as h->0. Does this limit look familiar? What does it define?
 
  • #3
But I suspect that this problem was intended to lead to the derivative of sin x and so that method would be inappropriate. What if you just try small values of h in
[tex]\frac{sin(2+h)- sin(2)}{h}[/tex] and see how small h has to be to so that you get the same answer to 2 decimal places?
 
  • #4
Halls, I actually did the equation like you have it there, and then using the sum rule for sine, sinacos b - cos asinb I got a final answer of cos 2 which equalled -0.42. I haven't plugged in any values to check my work, i just fiddled with the equation.
 
  • #5
All the problem asked is that you plug in values!

What I meant was, since sin(2)= 0.9093,
Taking h= 0.1, (f(2+h)- f(2))/h= (sin(2.1)- sin(2))/.1= -.4609.
Taking h= 0.01, (f(2+h)- f(2))/h= (sin(2.01)- sin(2))/.01= -.4206
Taking h= 0.001, (f(2+h)- f(2))/h= (sin(2.001)- sin(2))/.001= -.4166
Taking h= 0.0001, (f(2+h)- f(2))/h= (sin(2.0001)- sin(2))/.0001= -.4162
 
  • #6
I did it the way I did it because there was an example in my manual that led me to do it that way. I did it your way too and both ways led me to the same answer.

Thanks for your help!
 

1. What does it mean to evaluate a limit to 2 decimal places?

Evaluating a limit to 2 decimal places means finding the numerical value of a limit with a precision of two decimal places. It is a way to approximate the value of a limit and get a more accurate result than just using the limit notation.

2. Why is it important to evaluate a limit to 2 decimal places?

Evaluating a limit to 2 decimal places is important because it allows us to get a more precise understanding of the behavior of a function at a specific point. It can also help us make more accurate predictions and decisions in real-world applications, such as in physics and engineering.

3. How do you evaluate a limit to 2 decimal places?

To evaluate a limit to 2 decimal places, you can use a variety of methods such as direct substitution, factoring, or algebraic manipulation. You can also use numerical methods such as using a graphing calculator or using a table of values to approximate the limit.

4. Can a limit be evaluated to more or less than 2 decimal places?

Yes, a limit can be evaluated to more or less than 2 decimal places depending on the desired level of precision. In some cases, it may be necessary to evaluate a limit to a higher number of decimal places to get a more accurate result, while in other cases, 2 decimal places may be sufficient.

5. What are some potential sources of error when evaluating a limit to 2 decimal places?

When evaluating a limit to 2 decimal places, potential sources of error include rounding errors, incorrect use of mathematical methods, and using an inappropriate level of precision. It is important to double-check calculations and consider the context of the problem to avoid these errors.

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