$\lim _{x \rightarrow 0} \frac{(3^{2x}-\sqrt{x+1}) \sin 5x}{1-\cos 4x} =$

If $f(x) = \begin{cases} |x|+1, & x < 0 \\ 0, & x = 0 \\ |x|-1, & x > 0 \end{cases}$,for what value$(s)$ of $a$ does $\lim_{x \to a} f(x)$ exist?

$\lim _{x \rightarrow 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2}=$

Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ such that $\lim _{x \rightarrow \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin(1/x^2)}{x^{\alpha \beta}(\log_e(1+x))^\beta} = 0$. Then which of the following is (are) correct?

$\lim _{x \rightarrow 0}\left[\tan \left(x+\frac{\pi}{4}\right)\right]^{1 / x}$ is equal to

$\lim _{x \rightarrow 0} \frac{(1-e^x) \sin x}{x^2+x^3}$ is equal to