$\mathop {\lim }\limits_{x \to 0} \left( \frac{\sin x - x + \frac{x^3}{6}}{x^5} \right) = $

Let $f(x) = x^{6} + 2x^{4} + x^{3} + 2x + 3$,$x \in R$. Then the natural number $n$ for which $\lim_{x \rightarrow 1} \frac{x^{n} f(1) - f(x)}{x - 1} = 44$ is ...... .

If $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \infty$ and $\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k$,then $12 k=$

$\lim _{x \rightarrow 0} \left( \frac{x \cdot 10^x - x}{1 - \cos x} \right)$ is equal to

$\lim _{x \rightarrow \pi / 6} \left[ \frac{3 \sin x - \sqrt{3} \cos x}{6x - \pi} \right]$ is equal to:

$\mathop {\lim }\limits_{\theta \to \pi /2} (\sec \theta - \tan \theta ) = $