chris215
  • chris215
Find the exact value of the following limit: (e^(6x)-6x-1)/(x^2)
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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chris215
  • chris215
I got 18
SolomonZelman
  • SolomonZelman
x --> what ?
chris215
  • chris215
as x->0

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More answers

mathmale
  • mathmale
By (e^(6x)-6x-1)/(x^2) do you mean the following?\[\frac{ e ^{6x}-6x-1 }{ x^2 }?\]
SolomonZelman
  • SolomonZelman
\(\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~0} \frac{e^{6x}-6x-1}{x^2} }\)
chris215
  • chris215
yes
SolomonZelman
  • SolomonZelman
use L'Hospital's rule. (it is 0/0 now)
mathmale
  • mathmale
I agree with Solomon: Use l'H's Rule. Differentiate the numerator with respect to x. Differentiate the denominator with respect to x, separately. Form a fraction out of your results. Take the limit as x -> 0 of that fraction.
SolomonZelman
  • SolomonZelman
differentiate, and if you have questions, ask.
mathmale
  • mathmale
@chris215 : Share your thoughts, please, and especially share any work you've been able to do on this problem. Have you worked with l;Hopital's Rule before?
SolomonZelman
  • SolomonZelman
If you used L'Hospital's rule twice, then 18 is right
SolomonZelman
  • SolomonZelman
well, 18 is just correct...
chris215
  • chris215
I differentiated and got d/dx(e^(6 x)-6 x-1) = 6 (e^(6 x)-1)
chris215
  • chris215
and x^2= 2x
SolomonZelman
  • SolomonZelman
yes, very good \(\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~0} \frac{e^{6x}-6x-1}{x^2} }\) since it is not 0/0 when x=0 is plugged, we differentiate top and bottom. \(\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~0} \frac{\frac{d}{dx}\left[e^{6x}-6x-1\right] }{\frac{d}{dx}\left[x^2\right]} }\) \(\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~0} \frac{6e^{6x}-6}{2x} }\) \(\large\color{#000000}{\displaystyle\lim_{x \rightarrow ~0} \frac{3(e^{6x}-1)}{x} }\) 0/0 again, and thus differentiate again...
SolomonZelman
  • SolomonZelman
you will use the chain rule for the exponent again, and then evaluate the limit.... and you will get 18 - the correct result you already had.
SolomonZelman
  • SolomonZelman
((just asking, when you evaluated the limit first, you used an alternative approach... was it graphing? ))
Zarkon
  • Zarkon
taylor series is the quickest way
mathmale
  • mathmale
Please give @chris215 an opportunity to respond.
SolomonZelman
  • SolomonZelman
Zarkon, honestly, that is the first time I have ever heard this from anyone. To me (and to many others), finding L'H'S twice and differentiating is a matter of 1 minute...
chris215
  • chris215
yeah I used graphinng
SolomonZelman
  • SolomonZelman
but, whatever you say :)
SolomonZelman
  • SolomonZelman
Yeah, graphing is good, but not always available (eg tests)
chris215
  • chris215
true but thanks so much! that wa very helpful!
SolomonZelman
  • SolomonZelman
Anytime:) Just remember that if you see that when you plug in the variable into the limit you get 0/0 or ∞/∞ then you can differentiate top and bottom (this is L'Hospital's rule)....

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