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

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

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

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

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

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

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow ∞} \frac{ (1+7)^x }{ 12 }\]

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

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

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0In order to use L'hopitals rule, you must have a fraction with a function on the numerator and denominator

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

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

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

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

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

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

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