Some limits manipulation.

Feb 2020
5
0
Germany
Could someone explain the steps involved in this limit? I'm reading a proof and kinda stuck here.

For \(\displaystyle k \in \mathbb{Z}^+\) and \(\displaystyle c\) a positive constant, set \(\displaystyle \beta_k = c \sqrt{\frac{k}{\ln(k)}}\) and \(\displaystyle \delta_k = \frac{\ln(\frac{k}{\beta_k})}{k}\). Then \(\displaystyle \lim\limits_{k \to \infty} (\beta_k(1-\delta_k)^k + \beta_k^2\delta_k) = \frac{c^2}{2}\).
 
Dec 2015
972
128
Earth
\(\displaystyle l=\lim_{k\rightarrow \infty }[ \beta_k (1+\delta_{k^{-1}})^k +\beta_k^{2} \delta_k ] =\lim_{k\rightarrow \infty }[ \beta_k ( 1+\frac{\ln (\beta_k /k)}{k})^{k} +\beta_k^{2} \delta_k ] = \lim_{k\rightarrow \infty }[ \frac{\beta_k^{2}}{k} +c^2 \frac{k}{\ln(k)} \cdot \ln(\frac{k}{\beta_k })\frac{1}{k}] \).

\(\displaystyle l=\lim_{k\rightarrow \infty} \frac{c^2 }{\ln (k)} +c^2 \lim_{k \rightarrow \infty } \frac{\ln(\sqrt{k\ln(k)}) -\ln(c)}{\ln(k) }=\frac{c^2 }{2} \lim_{k\rightarrow \infty } \frac{\ln(k) + \ln \ln (k) }{\ln (k)}=\frac{c^2 }{2} +\frac{c^2 }{2} \lim_{k\rightarrow \infty } \frac{\ln \ln (k)}{\ln (k ) }=\frac{c^2 }{2} \).
 
Feb 2020
5
0
Germany
\(\displaystyle l=\lim_{k\rightarrow \infty }[ \beta_k (1+\delta_{k^{-1}})^k +\beta_k^{2} \delta_k ] =\lim_{k\rightarrow \infty }[ \beta_k ( 1+\frac{\ln (\beta_k /k)}{k})^{k} +\beta_k^{2} \delta_k ] = \lim_{k\rightarrow \infty }[ \frac{\beta_k^{2}}{k} +c^2 \frac{k}{\ln(k)} \cdot \ln(\frac{k}{\beta_k })\frac{1}{k}] \).

\(\displaystyle l=\lim_{k\rightarrow \infty} \frac{c^2 }{\ln (k)} +c^2 \lim_{k \rightarrow \infty } \frac{\ln(\sqrt{k\ln(k)}) -\ln(c)}{\ln(k) }=\frac{c^2 }{2} \lim_{k\rightarrow \infty } \frac{\ln(k) + \ln \ln (k) }{\ln (k)}=\frac{c^2 }{2} +\frac{c^2 }{2} \lim_{k\rightarrow \infty } \frac{\ln \ln (k)}{\ln (k ) }=\frac{c^2 }{2} \).
In the third equality, how do you get \(\displaystyle \lim\limits_{k \to \infty}\beta_k(1+ \frac{ln(\beta_k/k)}{k})^k = \lim\limits_{k \to \infty} \frac{\beta_k^2}{k}\)?
 

romsek

Math Team
Sep 2015
2,875
1,608
USA
$\lim \limits_{n\to \infty} \left(1 + \dfrac x n\right)^n = e^x$

$\lim \limits_{k\to \infty} \beta_k \left(1 + \dfrac{\ln\left(\frac{\beta_k}{k}\right)}{k}\right)^k = \\

\large \beta_k e^{\ln\left(\frac{\beta_k}{k}\right)} = \\

\beta_k \left(\dfrac{\beta_k}{k}\right) = \\

\dfrac{\beta_k^2}{k}

$
 
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