$\lim _{z \rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1} = $

The quadratic equation whose roots are $\ell = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^3 \theta}{\theta} \right)$ and $m = \lim_{\theta \rightarrow 0} \left( \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)} \right)$ is

If $L = \lim_{x^2 \to a} \frac{b - \cos(x^2 - a)}{(x^2 - a) \sin(c(x^2 - a))}$ is a non-zero finite value $(a > 0)$,then:

If $a, b, c$ and $k$ are non-zero real numbers and $\lim _{x \rightarrow \infty} x\left(a^{\frac{1}{x}}+b^{\frac{1}{x}}+c^{\frac{1}{x}}-3 k^{\frac{1}{x}}\right)=0$,then $k=$

$\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {1 - \cos 2(x - 1)} }}{{x - 1}}$

$\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{4}{{x - 1}}} \right)^{3x - 1}} = $