Why does the exponent n work when substituting in this e limit?

  • Thread starter Thread starter Nerpilis
  • Start date Start date
  • Tags Tags
    Limit
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
Nerpilis
Messages
19
Reaction score
0
ok I have this limit question that was done in class but i didn't catch it at the time but they grazed over a step where I'm not sure what the reasoning was.
[tex]\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n + 1} \right)^{n} = \lim_{n \rightarrow \infty} \left( 1 + \frac{ \frac{1}{n} }{ 1 + \frac{1}{n} } \right)^{n} = e[/tex]
I see the multiplication of one in the form of 1/n over 1/n and i know that [tex]\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^{n} = e[/tex] and i can see the similarity and possible substituions...but what happens to the 'n' exponet since it doesn't substitute nicely?
 
Last edited:
Physics news on Phys.org
I see no reason to divide the numerator and the denominator by n. You can simply use a substitution like m=n+1. Or write:
[tex]\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n + 1} \right)^{n} = \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n + 1} \right)^{n+1}\cdot \frac{1}{1+\frac{1}{n+1}}[/tex]
 
I think I'm a little more confused now...I do agree that i don't see what dividing by n did to help.