Let $p = \mathop {\lim }\limits_{x \to 0^+} (1 + \tan^2 \sqrt{x})^{\frac{1}{2x}}$,then $\log p = $ . . .

$\lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3} = $

The set of all values of $a$ for which $\lim_{x \rightarrow a}(\lfloor x-5 \rfloor - \lfloor 2x+2 \rfloor) = 0$,where $\lfloor \alpha \rfloor$ denotes the greatest integer less than or equal to $\alpha$,is equal to

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:

$\lim _{x \rightarrow \infty} \left(\frac{x+a}{x+b}\right)^{x}$ is equal to

Evaluate the limit: $\lim _{x}$ ${\rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)$