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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   -

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  • Submitted by: Galactus
  • on October 9, 2011
  • Category: Science And Technology
  • Length: 337 words
  • Views: 360
  • Popularity Rank: 362764
  • 1 rating(s)

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