$$\mathop {\lim }\limits_{x \to 0} {{\sqrt {1 - \cos 2x} } \over {\sqrt 2 x}}$$ is

Let $$f(2) = 4$$ and $$f'(x) = 4.$$

Then $$\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}$$ is given by

$$\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x}$$

$$\mathop {\lim }\limits_{x \to 0} {{\log {x^n} - \left[ x \right]} \over {\left[ x \right]}}$$, $$n \in N$$, ( [x] denotes the greatest integer less than or equal to x )