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

Let $f(x) = \int_0^x (t + \sin(1 - e^t)) dt, x \in R$. Then $\lim_{x \rightarrow 0} \frac{f(x)}{x^3}$ is equal to

$\lim _{x \rightarrow 0} \frac{\left(1+\frac{x}{2}\right)^{5 / 7}-1}{x} = $

If $\mathop {\lim }\limits_{x \to a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = - 1$,then

The limit of $\left\{\frac{1}{x} \sqrt{1+x}-\sqrt{1+\frac{1}{x^{2}}}\right\}$ as $x \rightarrow 0$ is:

Let $\alpha$ and $\beta$ be real numbers such that $\lim _{x \rightarrow 0} \frac{1}{x^3}\left(\frac{\alpha}{2} \int_0^x \frac{1}{1-t^2} d t+\beta x \cos x\right)=2$. Then the value of $\alpha+\beta$ is $....$ (in $.40$)