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Brooke_armyBest ResponseYou've already chosen the best response.1
I got the number 0.583
 one year ago

Brooke_armyBest ResponseYou've already chosen the best response.1
the correct answer is 1.79. I'm not sure where i went wrong
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
how did you get 0.583 ? mind showing your work/steps ?
 one year ago

mathmateBest ResponseYou've already chosen the best response.0
is it \( \large \frac{(1+7/x)^x}{12}\) or \( \large (1+7/x)^{x/12} \)
 one year ago

Brooke_armyBest ResponseYou've already chosen the best response.1
y=lim (1+7/x)^x/12 ln(y)=lim x/12 ln(1+7/x) \[\frac{ \ln(1+\frac{ 7 }{x} }{\frac{12 }x }\]
 one year ago

Brooke_armyBest ResponseYou've already chosen the best response.1
mathmate it's the second
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
\[\lim_{x \rightarrow ∞} \frac{ (1+7)^x }{ 12 }\]
 one year ago

Brooke_armyBest ResponseYou've already chosen the best response.1
then i took d/dx to both the numerator and the denominator
 one year ago

Brooke_armyBest ResponseYou've already chosen the best response.1
the problem is \[\lim_{x \rightarrow \infty} (1+\frac{ 7 }{ 12})^\frac{ x }{12 }\]
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
In order to use L'hopitals rule, you must have a fraction with a function on the numerator and denominator
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
do you know a general formula \[\lim_{y \rightarrow 0} (1+y)^{\frac{ 1}{y}}=...?\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
if you know ^ that, then you can put 7/x = y in your limit question first, to bring in that form.
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
if you use the formula, you get the correct answer in just few steps by adjusting the exponent. \[\lim_{y \rightarrow 0} (1+y)^{\frac{ 1}{y}}=e\]
 one year ago

Brooke_armyBest ResponseYou've already chosen the best response.1
im trying that right now
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
okay :) take your time...
 one year ago

Brooke_armyBest ResponseYou've already chosen the best response.1
thanks i got the right answer
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
good! you're welcome ^_^
 one year ago
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