Evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}$

Let $e$ denote the base of the natural logarithm. The value of the real number $a$ for which the right-hand limit $\lim_{x \rightarrow 0^{+}} \frac{(1-x)^{\frac{1}{x}}-e^{-1}}{x^a}$ is equal to a nonzero real number is:

$\lim _{x \rightarrow \frac{\pi}{2}} \frac{(1-\sin x)(8 x^3-\pi^3) \cos x}{(\pi-2 x)^4}$

$\mathop {\lim }\limits_{x \to 1} \frac{{x + {x^2} + ...... + {x^n} - n}}{{x - 1}}$ is equal to

$\lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$ is equal to

If $\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha$,then the value of $\frac{\log _e \alpha}{1+\log _e \alpha}$ equals :