Theory about limits

Dec 2015
868
116
Earth
Prove the following statements :
After the limit turns to be an equation where \(\displaystyle L=pL\) or another transformation , then :
For p>1 , \(\displaystyle L=\infty\) ; For p<1 \(\displaystyle L=0\)

Example: \(\displaystyle L=\lim_{n\rightarrow \infty} \dfrac{n^2 }{2^n }=\lim_{n\rightarrow \infty} \dfrac{n^2 +2n +1 }{2^{n+1} }=\dfrac{1}{2}L\).
Since \(\displaystyle p=1/2 < 1 \; \Rightarrow L=0\).
 
Jun 2019
455
237
USA
I don't see how that can be. If $L = pL$, then $L = p^{-1}L$, right?

$\displaystyle L=\lim_{n\rightarrow \infty} \frac{n^2 }{2^n }=\lim_{n\rightarrow \infty} \frac{n^2 -2n +1 }{2^{n-1} }=2L$
This fact doesn't change the limit, though.

Aren't $L=0$ and $L\rightarrow \infty$ both potential solutions to $L=pL$, regardless of $p$?


(Note: Probably playing a little fast and loose with the notation. I think the intended meaning is clear, though, right?)
 
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