I do not quite understand how you went from to and incorporated the term into the summation. Also how does the outlying disappear?
Assuming that all of this can be reduced down to would I now consider?
So here if you look at individual parts of the series you have: since would approach 0 as n approaches infinity. It would follow similarly for the denominator and would cause the whole series to converge to 0?
How does the fact that f(x) is differentiable at a point x=a and f'(a) = 1 come into play?
Let's write out the summation :
So, how many times are you subtracting ? Since there are 100 terms, you are subtracting it 100 times.
The outlying does not disappear. When you divide by a fraction, that is the same as multiplying by its reciprocal. So, inside the summation we have:
Now, you can factor from every term, bringing it outside the summation (you can do this since does not depend on ).
Use your limit laws. The limit of a sum is the sum of the limits, so long as those limits exist (and so long as the sum is finite... infinite sums require more work). So:
As , . So, we can rewrite this limit:
Next, you asked about how you use . Use the definition of the derivative. For a general function , you know (Hint: When you write it out, replace by )
For your question about the series converging to 0, no. You have the numerator and denominator approaching zero. 0/0 is not defined, so you need to simplify the expression before you can take the limit.
Ok I think I have this entirely put together, can you tell me if this looks right?
Now, since the limit of a sum is the sum of the limits, we can write:
As , . So, we can rewrite this limit:
Note: the definition of a derivative is
We have this here in our modified limit: .
So we can write
Is that straightforward and correct?