Evaluating a limit to 2 decimal places

[meeklobraca]

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?

 

[danago]

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?

 

[HallsofIvy]

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
\frac{sin(2+h)- sin(2)}{h} and see how small h has to be to so that you get the same answer to 2 decimal places?

 

[meeklobraca]

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.

 

[HallsofIvy]

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

 

[meeklobraca]

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!