What is the value of e when substituting n=infinity into the limit (1 + 1/n)^n?

  • Thread starter tahayassen
  • Start date
In summary: Rewrite the right side as [ln (1 + 1/n)]/(1/n).Take the limit of both sides. If necessary, use L'Hopital's...I took the log of both sides, and then I moved the n exponent to the coefficient, so I got:logy = nlog(1+1/n)But I'm not sure where to proceed from here.This is the right track. Now you have an indeterminate form of the type [0/0] so you can use L'Hopital's rule. Taking the derivative of the top and bottom you get:d/dx(log(1+
  • #36
Thanks! I understand now.
 
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  • #37
Actually, one last thing, if I look at your third term:

[tex]\frac { 1 }{ 2 } (n^{ 2 }-3n+2)n^{ n-2 }[/tex]

But shouldn't it be:

[tex]\frac { 1 }{ 2 } (n)(n-1)n^{ n-2 }\\ =\frac { 1 }{ 2 } ({ n }^{ 2 }-n)n^{ n-2 }[/tex]
 
  • #38
Yes, I tried doing the coefficients in my head when I wrote that and evidently made a mistake. Fortunately it didn't affect the leading term.
 

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