$\lim _{x \rightarrow \infty}\left(\frac{x+5}{x+2}\right)^{x+3}$ equals

$\mathop {\lim }\limits_{x \to 0} \frac{{x + 2\sin x}}{{\sqrt {{x^2} + 2\sin x + 1} - \sqrt {{{\sin }^2}x - x + 1} }}$ is

If $f(x) = \begin{cases} x & \text{if } x < 0 \\ 1 & \text{if } x = 0 \\ x^2 & \text{if } x > 0 \end{cases}$,then $\mathop {\lim }\limits_{x \to 0} f(x) = $

If $\alpha=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{1-\cos x}$ and $\beta=\lim _{x \rightarrow 0} \frac{x \cdot 2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}}$,then

The value of $\mathop {\lim }\limits_{x \to 0} \left[ \frac{\sqrt{a + x} - \sqrt{a - x}}{x} \right]$ is

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