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

$ABC$ is an isosceles triangle inscribed in a circle of radius $r$. If $AB = AC$ and $h$ is the altitude from $A$ to $BC$,and $P$ is the perimeter of $ABC$,then $\mathop {\lim }\limits_{h \to 0} \frac{\Delta }{{{P^3}}}$ equals (where $\Delta$ is the area of the triangle).

If $\alpha$ and $\beta$ are the roots of the equation $375x^2 - 25x - 2 = 0$,then $\lim_{n \to \infty} \sum_{r=1}^n \alpha^r + \lim_{n \to \infty} \sum_{r=1}^n \beta^r$ is equal to

$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals

Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right-hand limit $\lim_{x \rightarrow 0^{+}} \frac{(1-x)^{\frac{1}{x}}-e^{-1}}{x^a}$ is equal to a nonzero real number is:

$\lim_{x \rightarrow 0} \frac{x \left( e^{\frac{\sqrt{1+x^{2}+x^{4}}-1}{x}} - 1 \right)}{\sqrt{1+x^{2}+x^{4}}-1}$ is equal to: