$$\lim _\limits{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}=$$

If $$a=\lim _\limits{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}$$ and $$b=\lim _\limits{n \rightarrow \infty} \frac{1^2+2^2+3^2+\ldots+n^2}{n^3}$$, then

If $$f(x) = {{{4^{x - \pi }} + {4^{x - \pi }} - 2} \over {{{(x - \pi )}^2}}}$$, for $$x \ne \pi $$, is continuous at $$x=\pi$$, then k =

$$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} + 5x - 7} - x} \right) = $$