$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x + \log (1 - x)}}{{{x^2}}}$ is equal to

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:

$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{1}{x} - \frac{{\log (1 + x)}}{{{x^2}}}} \right] =$

$\lim _{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$ is equal to

The value of the limit $\mathop {\lim }\limits_{x \to 0} {\left\{ {{1^{\frac{1}{{{{\sin }^2}x}}}} + {2^{\frac{1}{{{{\sin }^2}x}}}} + \dots + {n^{\frac{1}{{{{\sin }^2}x}}}}} \right\}^{{{\sin }^2}x}}$ is

The value of the limit of $\frac{x^3 - x^2 - 18}{x - 3}$ as $x$ tends to $3$ is