L'Hospital's Rule

  • Thread starter naspek
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  • #1
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Find the following limit using L'Hospital rule..

lim..........(e^x) / [(e^x) + 7x]
x->infinity

= f'(x) / g'(x)
=e^x / (e^x) + 7
= infinity/infinity
=0
am i got it right?
 

Answers and Replies

  • #2
HallsofIvy
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Find the following limit using L'Hospital rule..

lim..........(e^x) / [(e^x) + 7x]
x->infinity

= f'(x) / g'(x)
=e^x / (e^x) + 7
Okay to here

= infinity/infinity
What does that mean?

=0
Why would you think that?

am i got it right?
One you have e^x/(e^x+ 7) either use L'Hospital's rule again or divide both numerator and denominator by e^x.
 
  • #3
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e^x/e^x = 1 ???
 
  • #4
CompuChip
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Err... wait. You said: when I take the limit directly I get infinity / infinity, which is not defined. So we need L'Hôpitals rule. I agree with that.
Then you take the derivatives, and get infinity / infinity again. Then suddenly, that is defined and equal to 0?

The idea is, that
[tex]\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} [/tex]
in this case, but on the right hand side you get something which is still of the form "infinity / infinity". So you will need to apply it again.

Afterwards, to check your answer, you could try multiplying numerator and denominator by exp(-x).
 

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