$\mathop {\lim }\limits_{n \to \infty } {({3^n} + {4^n})^{\frac{1}{n}}} = $

The value of $\mathop {Limit}\limits_{x \to \infty } \,\frac{{{{\left( {{2^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}\,\, - \,\,{{\left( {{3^{{x^n}}}} \right)}^{\frac{1}{{{e^x}}}}}}}{{{x^n}}}\,$ (where $n \in N$) is

If ${S_n} = \sum\limits_{k = 1}^n {{a_k}} $ and $\mathop {\lim }\limits_{n \to \infty } {a_n} = a,$ then $\mathop {\lim }\limits_{n \to \infty } \frac{{{S_{n + 1}} - {S_n}}}{{\sqrt {\sum\limits_{k = 1}^n k } }}$ is equal to

The value of $\mathop {\text{Limit}}\limits_{x \to \infty } \frac{{\cot ^{ - 1}}\left( {{x^{ - a}}\log _a x} \right)}{{\sec ^{ - 1}}\left( {{a^x}\log _x a} \right)}$ for $a > 1$ is equal to

$\lim_{x \rightarrow \frac{\pi}{2}} (\tan^{2} x (\sqrt{2 \sin^{2} x + 3 \sin x + 4} - \sqrt{\sin^{2} x + 6 \sin x + 2}))$ is equal to

The value of $\lim_{n \to \infty} \frac{(n + 2)! + (n + 1)!}{(n + 2)! - (n + 1)!}$ is