Prove that $ \lim_{x\to0} ({1^{(1/\sin^2x)} + 2^{(1/\sin^2x)} + 3^{(1/\sin^2x)} + ....+ n^{(1/\sin^2x)})^{\sin^2(x)}} = \frac{n(n+1)}{2} $ $$\lim_{x\to0} ({1^{(1/\sin^2x)} + 2^{(1/\sin^2x)} + 3^{(1/\sin^2x)} + ....+ n^{(1/\sin^2x)})^{\sin^2(x)}}  = \frac{n(n+1)}{2}$$
Attempt:
According to me, it has to be $1$, since the outermost exponent tends to $0$.
But anyways, would taking the limit as equal to $L$. And taking ($\log$) on both sides help?
 A: Let us consider $n=2$ first, for simplicity, and use $x^2$ instead of $\sin^2x$, as $x\to0$. As you suggested, since the function has the indeterminate form $\infty^0$ as $x\to0$, let us consider the logarithm of the original function
$$
x^2\log\left(1+2^{1/x^2}\right) = \frac{\log\left(1+2^{1/x^2}\right)}{1/x^2},
$$
which, as $x\to 0$, is of the form $\infty/\infty$. We can apply l'Hospital's rule getting
$$
\frac{\log 2}{1+2^{-1/x^2}},
$$
which tends to $\log 2$ as $x\to0$.
Therefore
$$
\lim_{x\to 0} \left(1+2^{1/x^2}\right)^{x^2}=2.
$$
In general, we may apply l'Hospital's rule to
$$
\frac{\log\left(1+2^{1/x^2}+3^{1/x^2}+\ldots+n^{1/x^2}\right)}{1/x^2}
$$
getting
$$
\frac{\left(\frac{2}{n}\right)^{1/x^2}\log 2+\left(\frac{3}{n}\right)^{1/x^2}\log 3+\ldots+\log n}{1+\ldots + \left(\frac{3}{n}\right)^{1/x^2}+\left(\frac{2}{n}\right)^{1/x^2}+n^{-1/x^2}},
$$
which tends to $\log n$ as $x\to0$, proiving
$$
\lim_{x\to0}\left(1+2^{1/x^2}+3^{1/x^2}+\ldots+n^{1/x^2}\right)^{x^2}=n.
$$
A: For any $n$ numbers $a_1, \ldots, a_n > 0$, we have
$$
\begin{align}
\lim_{p\to+\infty} (a_1^p + a_2^p + \ldots + a_n^p)^{1/p} 
&= \max\{ a_1, \ldots, a_n \}\tag{*1a}\\
\lim_{p\to-\infty} (a_1^p + a_2^p + \ldots + a_n^p)^{1/p} &= \min\{ a_1, \ldots, a_n \}\tag{*1b}
\end{align}$$
I will only justify $(*1a)$.
Relabeling the numbers if necessary, we only need to study the case $0 < a_1 \le a_2 \ldots \le a_n$.
For any $p > 0$, we have
$$ a_n = (a_n^p)^{1/p} \le (a_1^p + \ldots + a_n^p)^{1/p} \le (a_n^p + \ldots + a_n^p)^{1/p} = n^{1/p} a_n$$
Since $\lim\limits_{p\to+\infty} n^{1/p} = 1$, by squeezing, $(*1a)$ follows.
Apply this to the case $(a_1,\ldots,a_n) = (1,\ldots,n)$ and notice as $x \to 0$, $\frac{1}{\sin^2x} \to +\infty$, we get
$$\lim_{x\to 0} \left(1^{1/\sin^2 x} + \ldots + n^{1/\sin^2 x}\right)^{\sin^2 x}
= \max\{ 1, \ldots, n \} = n$$
