Question

If f(x) = sinx, evaluate f(2+h)-f(2)/h to two decimal places as h approaches 0. Help?!

Answer 1

The limit equals `-0.42` approximately.

Explanation:

We have

`L = lim_(h->0) (sin(2 + h) - sin(2))/h`

`L = lim_(h->0) (sin2cos h + cos2sin h - sin2)/h`

`L = lim_(h->0) (sin2cos h - sin2)/h +(cos2sin h)/h`

`L = lim_(h->0) sin2((cos h - 1)/h) + cos2(sin h)/h`

`L = lim_(h->0) sin2((cos h -1 )/h) + lim_(h->0) cos2(sin h)/h`

We now use the well known calculus limits of `lim_(theta ->0) (sin theta)/theta = 1` and `lim_(beta -> 0) (cosbeta - 1)/beta = 0` to evaluate.

`L = sin2(0) + cos2(1)`

`L = cos(2)`

Which can be approximated to

`L ~~ -0.42`

And I assumed that `cos(2)` was in radians.

Hopefully this helps!

Answer 2

`lim_(h->0)(f(2+h)-f(2))/h = f'(2) = cos(2) ~~ -0.42`

Explanation:

Alternatively, we can use the limit definition of the derivative:

`f'(x) = lim_(h->0)(f(x+h)-f(x))/h`

This exactly matches our limit if you plug in `x=2`, so what we're looking for then is:

`f'(2)`

Since `f(x)=sinx`, we know that the derivative of `f(x)` is:

`f'(x) = cosx`

Therefore our limit equals

`lim_(h->0)(f(2+h)-f(2))/h = f'(2) = cos(2) ~~ -0.42`

Final Answer