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1) f’(x) = limh->0 ( e2(x+h) - e2x ) / h = (1/h) limh->0 ( e2x ( e2h - 1) ) /h = e 2x limh->0 ((e2h -1) /h ) (The limit of a quotient is the quotient of the limits and the limit of a constant with respect to h is the constant.) . Let’s take the logarithim of the second half of the product. This is a perfectly valid operation since the logarithim is a continous function defined on all positive reals and the limit approaches 0 from the right. So the composition of the logarithim of a limit of a function defined on the positive reals will have the same limit as the original function. So:

limh->0 ((e2h -1) /h ) = lim In h->+0 ( In( e2h -1) - In h ) = lim In h->+ 0 ( 2h -0 - In h ) = 2 since In h ®0 as h ®1. So e 2x limh->0 ((e2h -1) /h ) = 2 e 2x . as expected.

(I’m not sure how you’d do this by an e-d limit argument, but this seems to work. Give me a good night’s sleep and 3 bucks more and I’ll figure it out for you………LOL )

2) We need to show the derivative limit exists at each case. First, the left hand side:

f(x) = { x2 if x < 1

2x - 1 if x £ 1

Since both x2 and 2x-1 are continuous on the domain at all points other then 1. f (x) is continuous since lim- x->1 f(x) =1 and lim+ x->1 f(x) =1. (Not hard to prove,we’ll prove the left hand limit and leave the other as an exercise.)

÷ f(x) -1 ÷ = ú x2 -1 ú = ú x +1 ú ú x -1 ú . Let d= e/ ú x +1 ú . Then

÷f(x) -1 ÷ = ú x +1 ú ú x -1 ú < (e/ ú x +1 ú ) ú x +1 ú = e.

Now lastly, we need to prove the limit that defines the derivative exists on the whole domain of the step function. For x < 1:

Lim h->0(f (x+h) - f(x) )/h = Lim h->0 ( (x+h)2 -