If $$\alpha, \beta$$ are the roots of the equation $$a x^2+b x+c=0$$ then $$\lim _\limits{x \rightarrow \beta} \frac{1-\cos \left(a x^2+b x+c\right)}{(x-\beta)^2}$$ is

$$\mathop {\lim }\limits_{x \to \infty } \left\{ {x - \root n \of {(x - {a_1})(x - {a_2})\,...\,(x - {a_n})} } \right\}$$ where $${a_1},{a_2},\,...,\,{a_n}$$ are positive rational numbers. The limit

Let $$f:[1,3] \to R$$ be continuous and be derivable in (1, 3) and $$f'(x) = {[f(x)]^2} + 4\forall x \in (1,3)$$. Then

f(x) is a differentiable function and given $$f'(2) = 6$$ and $$f'(1) = 4$$, then $$L = \mathop {\lim }\limits_{h \to 0} {{f(2 + 2h + {h^2}) - f(2)} \over {f(1 + h - {h^2}) - f(1)}}$$