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

If $[x]$ denotes the greatest integer less than or equal to $x$,then the value of $\mathop {\lim }\limits_{x \to 1} (1 - x + [x - 1] + [1 - x])$ is

If $a, b, c$ and $k$ are non-zero real numbers and $\lim _{x \rightarrow \infty} x\left(a^{\frac{1}{x}}+b^{\frac{1}{x}}+c^{\frac{1}{x}}-3 k^{\frac{1}{x}}\right)=0$,then $k=$

The quadratic equation whose roots are $\ell = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^3 \theta}{\theta} \right)$ and $m = \lim_{\theta \rightarrow 0} \left( \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)} \right)$ is

$\lim _{n \rightarrow \infty} \frac{\left[6^2+12^2+18^2+\ldots+(6 n)^2\right]^2}{[5+10+15+\ldots+5 n]\left[2^3+4^3+6^3+\ldots+(2 n)^3\right]} =$

$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$ is equal to