If $a = \lim_{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2}}{x^4}$ and $b = \lim_{x \rightarrow 0} \frac{\sin^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$,then the value of $ab^3$ is

The value of $\mathop {\lim }\limits_{n \to \infty } {\left( {e \cdot {a^2} \cdot {e^3} \cdot {a^4} \cdots {e^{n - 1}} \cdot {a^n}} \right)^{\frac{1}{{{n^2} + 1}}}}$ is equal to

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

$\lim _{n \rightarrow \infty} \frac{1}{n^3+1}+\frac{4}{n^3+1}+\frac{9}{n^3+1}+\ldots+\frac{n^2}{n^3+1} = $

$\mathop {\lim }\limits_{x \to 0^ + } \frac{x e^{1/x}}{1 + e^{1/x}} = $

$\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 }}{{{y^4}}} = $