Taking the derivatives you get { [1/(1+x)] -1}/-2in(2x).
Now simplify and find the limits.
Use L'opital to find the following limit:
Second Q (unrelated):
Use the definition of a derivative (which I know) to show that cos x is the derivative of sin x.
^ Ok for this I wrote it out with the h's as per normal. I figured, Change the numerator using sina-sinb = 2sin(x-y/2)cos(x+y/2) but then I'm a little brammed.
For the first q, the l'Opital, I make it zero but I'm having trouble convincing myself. Both the derivatives are 0 at x=0?
Is it not ok to stick with what I get? Or does L'Hopital HAVE to result in a none-zero denominator? Is that the only reason to continue? I am convinced that f(x), g(x), f'(x) AND g'(x) are all zero at x=0...
How about this question@
Show cos x is the derivative of sin x using the definition of a derivative
The 'fundamental limit' has to be prooved 'geometrically' as in...
Limits Of Trigonometric Functions | Limits Theorems | Tutorvista.com
Kind regards