Let $f: R \rightarrow R$ be a function such that $f(2)=4$ and $f^{\prime}(2)=1$. Then,the value of $\lim _{x \rightarrow 2} \frac{x^{2} f(2)-4 f(x)}{x-2}$ is equal to:

$\lim _{x \rightarrow 0} \frac{e^{\tan x}-e^x}{\tan x-x} = $

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

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

Evaluate the limit: $\mathop {\text{Limit}}\limits_{x \to 4} \frac{(\cos \alpha)^x - (\sin \alpha)^x - \cos 2\alpha}{x - 4}$,where $0 < \alpha < \frac{\pi}{2}$.

Let $[x]$ denote the greatest integer less than or equal to $x$ and $k \geq 2$ be an integer. Then $\lim_{x \rightarrow k} \frac{\sin \left(2 \pi\left([x]-\left[\frac{x}{k}\right]\right)-x\right)+\sin k}{x-k} = $