$\lim _{x \rightarrow 8} \frac{\sqrt{1+\sqrt{1+x}}-2}{x-8}$ is equal to

If $L = \lim_{x^2 \to a} \frac{b - \cos(x^2 - a)}{(x^2 - a) \sin(c(x^2 - a))}$ is a non-zero finite value $(a > 0)$,then:

Let $f(x) = 5 - |x - 2|$ and $g(x) = |x + 1|$,where $x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim_{x \to \alpha \beta} \frac{(x - 1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$ is equal to:

If $\alpha = \lim_{x \rightarrow 0^{+}} \left( \frac{e^{\sqrt{\tan x}} - e^{\sqrt{x}}}{\sqrt{\tan x} - \sqrt{x}} \right)$ and $\beta = \lim_{x \rightarrow 0} (1 + \sin x)^{\frac{1}{2} \cot x}$ are the roots of the quadratic equation $ax^2 + bx - \sqrt{e} = 0$,then $12 \log_e(a + b)$ is equal to.............

Evaluate $\mathop {\lim }\limits_{x \to 0} f(x),$ where $f(x) = \begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x=0 \end{cases}$

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