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Finding limit using l'Hopitals rule

  1. Mar 29, 2012 #1
    Hello, I am tying to use l'Hopital's rule to solve this limit:
    {e^(5+h)-e^5} / h
    limit h tending towards 0

    Using l'Hopitals rule I differentiate both numerator and denominator to get:
    e^(5+h)-e^5 / 1
    THen plugging 0 back in I get 0/1 which would give me a limit of 0 ?
    But I think the limit should actually be e^5.

    Can someone see where I have gone wrong ?
    Thanks kindly
     
    Last edited: Mar 29, 2012
  2. jcsd
  3. Mar 29, 2012 #2
    What is the rate of change of e^5 with respect to h? I am assuming you are dealing with { e(5+h) - e^5 }/h.
     
    Last edited: Mar 29, 2012
  4. Mar 29, 2012 #3
    yes, that is correct. I am trying to apply l'Hopital's rule to that formula to obtain the limit as h tends towards 0.
    I dont think I have it right in my attempt above. Any help would be appreciated.

    Thank you.
     
  5. Mar 29, 2012 #4

    Curious3141

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    e^5 is a constant. What's the derivative of a constant?
     
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