If $f(x)$ is continuous and $f\left( \frac{9}{2} \right) = \frac{2}{9}$,then $\lim_{x \to 0} f \left( \frac{1 - \cos 3x}{x^2} \right)$ is equal to:

Let $f(x) = \begin{cases} \frac{2^x + 2^{3-x} - 6}{\sqrt{2^{-x}} - 2^{1-x}} & \text{if } x > 2 \\ \frac{x^2 - 4}{x - \sqrt{3x - 2}} & \text{if } x < 2 \end{cases}$. Determine the nature of the function at $x = 2$.

Let $f(x) = \begin{cases} \frac{x^4-5x^2+4}{|(x-1)(x-2)|} & , x \neq 1,2 \\ 6 & , x=1 \\ 12 & , x=2 \end{cases}$. Then $f(x)$ is continuous on the set:

If $f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ 3, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then the value of $k$ is equal to . . . . . . .

If $f(x) = \frac{1+\cos \pi x}{\pi(1-x)^2}$ for $x \neq 1$ is continuous at $x=1$,then $f(1)$ is equal to

Show that every polynomial function is continuous.