If $$f(x)=\left\{\begin{array}{cc}\frac{1-\cos K x}{x \sin x}, & \text { if } x \neq 0 \\ \frac{1}{2}, & \text { if } x=0\end{array}\right.$$ is continuous at $$x=0$$, then the value of $$K$$ is

If $$f(x)=\left\{\begin{array}{cl}\frac{\sin 3 x}{e^{2 x}-1} ; & x \neq 0 \\ k-2 ; & x=0\end{array}\right.$$ is continuous at $$x=0$$, then $$k=$$

$$\sum_\limits{r=1}^n(2 r-1)=x$$ then, $$ \lim _\limits{n \rightarrow \infty}\left[\frac{1^3}{x^2}+\frac{2^3}{x^2}+\frac{3^3}{x^2}+\ldots+\frac{n^3}{x^2}\right]=$$

Rolle's theorem is not applicable in which one of the following cases?