Evaluate the given limit: $\mathop {\lim }\limits_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2}$

If $\alpha, \beta$ are the roots of the quadratic equation $ax^2 + bx + c = 0$,then $\lim_{x \to \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}$ equals

If $\Delta(x) = \begin{vmatrix} e^x & -1 \\ \sin x - 1 & 1 \end{vmatrix}$,then $\lim_{x \rightarrow 0} \frac{\Delta(x)}{x} = $

Evaluate: $\cos \left[ \lim_{x \rightarrow \infty} \frac{2 \pi |x| + \pi x}{|x| - 3x} + \lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} \cos^2 x \right)}{x^2} \right]$

If $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x} - \frac{4}{{{x^2}}}} \right)^{2x}} = {e^3},$ then $a$ is equal to

$\lim _{x \rightarrow \infty}\left(\frac{x+6}{x+1}\right)^{x+4}$ is equal to