Evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}$

$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum\limits_{r = 0}^n {{{\tan }^{ - 1}}\left( {1 + r + {r^2}} \right)} }}{n}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{x + 2\sin x}}{{\sqrt {{x^2} + 2\sin x + 1} - \sqrt {{{\sin }^2}x - x + 1} }}$ is

Let $f(x) = 5 - |x - 2|$ and $g(x) = |x + 1|$,where $x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim_{x \to \alpha \beta} \frac{(x - 1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$ is equal to:

The value of $\lim_{x \to 3} \frac{[x] - 3}{x - 3}$, where $[\cdot]$ denotes the greatest integer function, is...

If $\alpha$ and $\beta$ are the roots of the equation $375x^2 - 25x - 2 = 0$,then $\lim_{n \to \infty} \sum_{r=1}^n \alpha^r + \lim_{n \to \infty} \sum_{r=1}^n \beta^r$ is equal to