A towering limit

Relevant wiki: Limits by Logarithms

Let the limit be $y$ , then

$\begin{aligned} \ln y &= &\lim_{n\to\infty}\left( 1 + \dfrac \pi n\right)^n \ln \left( n \sin \dfrac \pi n \right) \\ \ln (\ln y) &=&\lim_{n\to\infty} n \ln \left( 1 + \dfrac \pi n \right) +\ln \left[ \ln \left( n \sin \dfrac \pi n \right) \right ] \\ \end{aligned}$

Applying the standard limits, we get

$\begin{aligned} \ln ( \ln y) &=& \pi + \ln ( \ln \pi ) \\ &=& \ln (e^\pi) + \ln \ln \pi \\ &=& \ln \left [ (e^\pi) \cdot \ln \pi\right ] \\ &=& \ln \left [ \ln \left( \pi^{e^\pi} \right)\right ] \\ \end{aligned}$

Hence, $\ln ( \ln y) = \ln \left [ \ln \left( \pi^{e^\pi} \right)\right ]$ , or $y = \pi^{e^\pi}$ . Hence, the limit is equal to $\pi^{e^\pi}$ .

Or the easiest we can do is to find the limit of (1+Π\n)^n separately that comes to be e^Π. And then the evaluation of (n[sin{Π\n}]) which comes to be Π . and after taking out limits of both the function we can combine our result and get Π^e^Π

Taking the natural log on both sides -

$\ln(L) =\lim_{n \to \infty}(1+\frac{\pi}{n})^n \cdot (n\sin \frac {\pi}{n}) = e^{\pi} \lim_{n \to \infty} \ln(n\sin \frac {\pi}{n})$

Let $\lim_{n \to \infty} \ln(n\sin \frac {\pi}{n}) = L_1$ . Using L'Hopital's

$\Rightarrow e^{L_1} = \lim_{n \to \infty} \dfrac {\sin(\frac {\pi}{n})}{\frac {1}{n}} = \lim_{n \to \infty} \dfrac {\frac {-\pi}{n^2} \cdot \cos(\frac {\pi}{n})}{\frac {-1}{n^2}} = \pi \cdot \lim_{n \to \infty} \cos\left(\frac {\pi}{n}\right) = \pi$

$\Rightarrow L_1 = \ln(\pi)$

Therefore

$L = \large e^{e^{\pi} \cdot \ln(\pi)} = \large \pi^{e^{\pi}} \approx \boxed {\large 3.19 \times 10^{11}}$

| The limits are both finite in the division!