Let $f(x) = \int_0^x (t + \sin(1 - e^t)) dt, x \in R$. Then $\lim_{x \rightarrow 0} \frac{f(x)}{x^3}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x}}{{{x^3}}}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{{{\sin }^{ - 1}}x}} = $

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} \frac{{{e^{\sin x}} - 1}}{x} = $

If $f: R \rightarrow R$ is such that $f(3)=16$ and $f^{\prime}(3)=4$,then find the value of $\lim _{x \rightarrow 3} \frac{x f(3)-3 f(x)}{x-3}$.