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What is Lim x tends to 0 (1 + kx) to the power of 1/kx ?

What is Lim x tends to 0 (1 + kx) to the power of 1/kx ?
 

Grade:12

1 Answers

Nandana
110 Points
7 years ago
hi,
 lim x → 0 (1+kx)^(1/kx) is given by expanding the Binomial series 
    limx → 0 {(01/kx)(kx)^0 +(11/kx)(kx)^1+(21/kx)(kx)^2+ …. +(1/kx1/kx) (kx)^1/kx) }
          =lim x → 0  { 1+ (kx)/1! +1/2!(1- kx) + 1/3! (1+2*(kx)^2 -3*kx ) + … }
          =  1+ 1/1!  + ½! + 1/3! + …
         = e (2.718281828 ….. )
 

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