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

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

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

mathmate
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1mathmate it's the second

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

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

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

abb0t
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1im trying that right now

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

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

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