If $\mathop {\lim }\limits_{x \to 0} \phi (x) = {a^3}, (a \ne 0)$; then $\mathop {\lim }\limits_{x \to 0} \phi \left( {\frac{x}{a}} \right)$ is equal to :-

$\lim _{x \rightarrow \infty} \frac{(2x^2-3x+5)(3x-1)^{x/2}}{(3x^2+5x+4)\sqrt{(3x+2)^x}}$ is equal to:

The quadratic equation whose roots are $l$ and $m$,where
$\begin{aligned}
& l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right), \\
& m=\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}, \text{ is}
\end{aligned}$

$\mathop {\lim }\limits_{x \to a} \frac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }} = \dots$ (where $a \ne 0$)

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^3} - 1}}{{{x^2} + 5x - 6}} = $

$\mathop {\lim }\limits_{x \to 0} \left( {\frac{{{a^x} - {b^x}}}{x}} \right) = $