find the value
lim [sin(x+h)-sinx]/h
h>0
There are many proofs of this, such as the one here:
http://www.wyzant.com/resources/lessons/math/calculus/derivative_proofs/sinx
However, it relies on the fact that sinx/x -> 1 as x->0. That is usually proven as given here:
http://math.ucsd.edu/~wgarner/math20a/sin%28x%29_over_x.htm
To find the value of the given expression, we need to evaluate the limit as h approaches 0. Here's how to do it:
1. Start with the expression: lim [sin(x+h) - sin(x)] / h as h approaches 0.
2. Notice that we have a difference of sine functions in the numerator. We can use the sine difference identity: sin(a) - sin(b) = 2 * cos((a + b) / 2) * sin((a - b) / 2). Applying this identity, we rewrite the expression as:
lim [2 * cos((x + h + x) / 2) * sin((x + h - x) / 2)] / h as h approaches 0.
3. Simplify the expression:
lim [2 * cos((2x + h) / 2) * sin(h / 2)] / h as h approaches 0.
4. Further simplify:
lim [2 * cos(x + h/2) * sin(h / 2)] / h as h approaches 0.
5. Apply the limit properties:
lim(2) * lim(cos(x + h/2)) * lim(sin(h / 2)) / lim(h) as h approaches 0.
We can simplify this to:
2 * cos(x) * sin(0) / 0 as h approaches 0.
6. Evaluate the limits:
lim(cos(x + h/2)) = cos(x), as h approaches 0. (Since cosine is a continuous function)
lim(sin(h / 2)) = sin(0) = 0, as h approaches 0.
lim(h) = 0, as h approaches 0.
7. Substitute the limits:
2 * cos(x) * sin(0) / 0 = 2 * cos(x) * 0 / 0.
8. Simplify:
2 * 0 / 0 = 0.
Therefore, the value of the expression lim [sin(x+h) - sin(x)] / h as h approaches 0 is 0.