$\mathop {\lim }\limits_{\theta \to {0^ + }} {(\sin \theta )^{(\sin \theta - {{\sin }^2}\theta )}}$ is

$\mathop {\lim }\limits_{x \to a} \frac{{\cos x - \cos a}}{{\cot x - \cot a}} = $

Let $f(1) = g(1) = k$ and their $n^{th}$ derivatives $f^{(n)}(1), g^{(n)}(1)$ exist and are not equal for some $n$. If $\lim _{x \rightarrow 1} \frac{f(1) g(x) - f(1) - g(1) f(x) + g(1)}{g(x) - f(x)} = 4$,then the value of $k$ is:

$\mathop {\lim }\limits_{x \to 0} {\left\{ {\tan \left( {\frac{\pi }{4} + x} \right)} \right\}^{1/x}} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{\log _e}(1 + x)}{{3^x - 1}} = $

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