Question

Let $a_n(n\ge 1)$ be the value of $x$ for which $\underset{x}{\overset{2x}{\int }}{e}^{-t^n}dt(x>0)$ is maximum. If $L=\underset{n\to \infty }{\mathrm{Lim}}\ln \left(a_n\right)$ then find the value of $e^{-L}$.

Answer 1

Answer: 2

Let $f_n(x)=\underset{x}{\overset{2x}{\int }}{e}^{-t^2}dt$
$f_n^{\prime }(x)=2\cdot e^{-(2x{)}^n}-e^{-x^n}$
For maxima and minima, $f_n^{\prime }(x)=0\Rightarrow 2e^{-(2x{)}^n}=e^{-x^n}\Rightarrow 2\cdot e^{-\left(2^n{x}^n\right)}=e^{-x^n}$ Taking log on both sides, we get
$\ln 2-2^n{x}^n=-x^n\Rightarrow \ln 2=x^n\left(2^n-1\right)\Rightarrow x^n=\frac{\ln 2}{2^n-1}\Rightarrow x={\left(\frac{\ln 2}{2^n-1}\right)}^{\frac{1}{n}}=a_n$
Also, $f_n^{\prime \prime }\left(x={\left(\frac{\ln 2}{2^n-1}\right)}^{\frac{1}{n}}\right)<0\Rightarrow f_n(x)$ is maximum at $x={\left(\frac{\ln 2}{2^n-1}\right)}^{\frac{1}{n}}$.
Now, $\ln a_n=\frac{\ln \left(\frac{\ln 2}{2^n-1}\right)}{n}=\frac{\ln (\ln 2)-\ln \left(2^n-1\right)}{n}$
Hence $L=\underset{n\to \infty }{\mathrm{Lim}}\ln \left(a_n\right)=\underset{n\to \infty }{\mathrm{Lim}}\frac{\ln (\ln 2)-\ln \left(2^n-1\right)}{n}=\underset{n\to \infty }{\mathrm{Lim}}\left(\frac{\ln (\ln 2)}{n}-\frac{\ln \left(2^n\left(1-\frac{1}{2^n}\right)\right)}{n}\right)$
$L=\underset{n\to \infty }{\mathrm{Lim}}\left(0-\frac{n\cdot \ln 2+\ln \left(1-\frac{1}{2^n}\right)}{n}\right)=-\ln 2$
Hence, $e^{-L}=2$