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luo
 2 years ago
Best ResponseYou've already chosen the best response.0Begin with \[\lim_{n \rightarrow \infty} \ln(1 + \frac{ 1 }{ n } ) ^{n} \] \[= \lim_{n \rightarrow \infty} n . \ln (1 + \frac{ 1 }{ n } ) \] \[= \lim_{n \rightarrow \infty } \frac{ \ln( 1+ \frac{ 1 }{ n }) }{ \frac{ 1 }{ n } } \] \[= \lim_{k \rightarrow 0} \frac{ \ln( 1 + k) }{ k } \](when n goes to infinity ,k = 1/n goes to zero ) \[= \lim_{k \rightarrow 0} \frac{ \ln(1 + k)  \ln(1) }{ k }\] ( ln(1) = 0 ) \[= ( \ln x) \prime(when,x = 1) \] \[= \frac{ 1 }{ x } (when, x = 1) \] \[= 1\] \[\lim_{n \rightarrow \infty} \ln(1 + \frac{ 1 }{ n } ) ^{n} = 1\] \[e ^{\lim_{n \rightarrow \infty} \ln(1 + \frac{ 1 }{ n } ) ^{n}} = e ^{1}\] then get, \[\lim_{n \rightarrow \infty} (1 + \frac{ 1 }{ n } ) ^{n} = e\] Proofed !
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