$\lim _{x \rightarrow 1} \frac{\tan \left(x^{2}-1\right)}{x-1}$ is equal to

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

$\mathop {\lim }\limits_{x \to 0^+} {x^x} = $

Value of $\mathop {\lim }\limits_{x \to 1 } \frac{{\left( {\log \left( {1 + x} \right) - \log 2} \right)\left( {3 \cdot 4^{x - 1} - 3x} \right)}}{{\left( {{{\left( {7 + x} \right)}^{1/3}} - {{\left( {1 + 3x} \right)}^{1/2}}} \right)\sin \pi x}}$

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

Let $g(x) = \frac{(x-1)^n}{\log \cos^m(x-1)}$ for $x \neq 1$,and let $p$ be the left-hand derivative of $|x-1|$ at $x=1$. If $\lim_{x \rightarrow 1^{+}} g(x) = p$,then: