$$ \lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n\left(k^2 x\right)= $$

The quadratic equation whose roots are

$$ l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^3 \theta}{\theta}\right) \text { and } m=\lim _{\theta \rightarrow 0}\left(\frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}\right) \text { is } $$

On differentiation if we get $f(x, y) d y-g(x, y) d x=0$ from $2 x^2-3 x y+y^2+x+2 y-8=0$, then $\frac{g(2,2)}{f(1,1)}=$

If $f(x)=e^x, h(x)=(f \circ f)(x)$, then $\frac{h^{\prime}(x)}{h(x)}=$