If $0 < x < y$,then $\mathop {\lim }\limits_{n \to \infty } {({y^n} + {x^n})^{1/n}}$ is equal to

$\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}$

If $l = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^3 \theta}{\theta} \right)$ and $m = \lim_{\theta \rightarrow 0} \left( \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)} \right)$,find the quadratic equation whose roots are $l$ and $m$.

$\lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2}$ is

Find $\mathop {\lim }\limits_{x \to 1} f(x)$,where $f(x) = \begin{cases} x^{2}-1, & x \leq 1 \\ -x-1, & x > 1 \end{cases}$

$\mathop {\lim }\limits_{x \to 0} \frac{{\tan x - \sin x}}{{{x^3}}} = $