Find the values of $k$ so that the function $f$ is continuous at the indicated point. $f(x) = \begin{cases} kx^2, & \text{if } x \le 2 \\ 3, & \text{if } x > 2 \end{cases}$ at $x=2$.

If $f(x) = \lim_{n \to \infty} \frac{[x^2] + [(2x)^2] + [(3x)^2] + \cdots + [(nx)^2]}{n^3}$,then $f(x)$ is (Where $[\cdot]$ is the Greatest Integer Function).

Let $R$ be the set of all real numbers and $\alpha \in R$ be positive. Define a function $f: R \rightarrow R$ by $f(0)=0$ and $f(x)=|x|^\alpha \sum \limits_{n=0}^{\infty}\left(1+x^2\right)^{-n}$,for $x \neq 0$. Then the set of real numbers $\alpha$ for which $f$ is continuous at $x = 0$ has

Is the function $f$ defined by $f(x) = \begin{cases} x, & \text{if } x \le 1 \\ 5, & \text{if } x > 1 \end{cases}$ continuous at $x=0$? At $x=1$? At $x=2$?

$A$ function is $f(x) = \begin{cases} \frac{e^{1/x}-1}{e^{1/x}+1}, & \text{if } x \neq 0 \\ 0, & \text{if } x=0 \end{cases}$

If $f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ \frac{1}{2}, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then the value of $k$ is . . . . . . .