Consider the function defined by
Find values for the constants and such that the derivative exists.
We know that
exists if and only if
exists. This limit exists if and only if the two one-sided limits exits and are equal:
Using the definition of , we then evaluate these limits,
In simplifying the right hand side we used that . Furthermore, for the limit on the left to exist we must have (otherwise the limit will diverge as ). Now for the expression on the right, we claim the limit in the expression is 0. We can see this because
But this limit is the derivative of at . Since and , this term is indeed 0.
So, coming back to our original equations we then have,
Furthermore, since we already established that we have,
Therefore, the expressions for and we are asked to find are,
Consider the function for a positive integer . Prove that
Conclude that by considering the limit of this as .
Proof. We recall the binomial theorem,
So, then we calculate,
Taking the limit as of both sides of the equation we conclude,
First, we simplify the expression,
Then, we evaluate the limit,
Proof. We use the sine and cosine of sum and difference formulas in the following computations: