If $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is

If $f(x)$,defined below,is continuous at $x = 4$,then find the values of $a$ and $b$ given that $f(x)$ is continuous on the interval $[0, 8]$.
$f(x) = \begin{cases} x^2 + ax + b, & 0 \leq x < 2 \\ 3x + 2, & 2 \leq x \leq 4 \\ 2ax + 5b, & 4 < x \leq 8 \end{cases}$

Let $f:(0,1) \rightarrow R$ be a function defined as $f(x) = \sqrt{n}$ if $x \in \left[\frac{1}{n+1}, \frac{1}{n}\right)$ where $n \in N$. Let $g:(0,1) \rightarrow R$ be a function such that $\int_{x^2}^x \sqrt{\frac{1-t}{t}} dt < g(x) < 2\sqrt{x}$ for all $x \in (0,1)$. Then $\lim_{x \rightarrow 0} f(x)g(x)$

Let $f: R \rightarrow R$ be a differentiable function such that $f(3)=3$ and $f^{\prime}(3)=\frac{1}{27}$. If $g(x)=\begin{cases} \int_3^{f(x)} \frac{3t^2}{x-3} dt & \text{if } x \neq 3 \\ K & \text{if } x=3 \end{cases}$ is continuous at $x=3$,then $K=$

In order that the function $f(x) = (x + 1)^{1/x}$ is continuous at $x = 0$,$f(0)$ must be defined as

If a function $f(x) = \begin{cases} \frac{\sqrt[3]{1+ax^2+bx^3}-\sqrt[3]{1-ax^2-bx^3}}{x^2}, & x < 0 \\ 5, & x=0 \\ \frac{\tan 3x - \sin 3x}{bx^3}, & x > 0 \end{cases}$ is continuous at $x=0$,then the geometric mean of $a$ and $b$ is