$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}} = $

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

Let $x_0$ be the real number such that $e^{x_0}+x_0=0$. For a given real number $\alpha$,define $g(x)=\frac{3 x e^x+3 x-\alpha e^x-\alpha x}{3\left(e^x+1\right)}$ for all real numbers $x$. Then which one of the following statements is True?

$\lim _{x \rightarrow 1} \frac{\log x}{1-x} = $

Let $\alpha$ be a positive real number. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $g: (\alpha, \infty) \rightarrow \mathbb{R}$ be the functions defined by $f(x) = \sin \left(\frac{\pi x}{12}\right)$ and $g(x) = \frac{2 \log_{e}(\sqrt{x}-\sqrt{\alpha})}{\log_{e}(e^{\sqrt{x}}-e^{\sqrt{\alpha}})}$. Then the value of $\lim_{x \rightarrow \alpha^{+}} f(g(x))$ is

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