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Finding limit using l'Hopitals rule

by fran1942
Tags: lhopitals, limit, rule
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fran1942
#1
Mar29-12, 02:26 AM
P: 80
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
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RoshanBBQ
#2
Mar29-12, 03:06 AM
P: 280
Quote Quote by fran1942 View Post
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
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.
fran1942
#3
Mar29-12, 03:40 AM
P: 80
Quote Quote by RoshanBBQ View Post
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.
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.

Curious3141
#4
Mar29-12, 03:57 AM
HW Helper
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P: 2,952
Finding limit using l'Hopitals rule

Quote Quote by fran1942 View Post
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
e^5 is a constant. What's the derivative of a constant?


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