If $\lim _{x \rightarrow \infty}\left(\left(\frac{e}{1-e}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x=\alpha$,then the value of $\frac{\log _e \alpha}{1+\log _e \alpha}$ equals :

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

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{(2x + 1)}^{40}}{{(4x - 1)}^5}}}{{{{(2x + 3)}^{45}}}} = $

$\mathop {\lim }\limits_{n \to \infty } \frac{{[{1^2}x + {1^2}] + [{2^2}x + {2^2}] + [{3^2}x + {3^2}] + \dots + [{n^2}x + {n^2}]}}{{{n^3}}}$ is equal to :- (where $[.]$ denotes the greatest integer function)

The value of $\lim _{x \rightarrow 0^{+}} \frac{\cos ^{-1}\left(x-[x]^{2}\right) \cdot \sin ^{-1}\left(x-[x]^{2}\right)}{x-x^{3}},$ where $[x]$ denotes the greatest integer $\leq x$ is

Let $a_{1}, a_{2}, \dots, a_{n}$ be fixed real numbers and define a function $f(x) = (x - a_{1})(x - a_{2}) \dots (x - a_{n})$. What is $\lim_{x \to a_{1}} f(x)$? For some $a \neq a_{1}, a_{2}, \dots, a_{n}$,compute $\lim_{x \to a} f(x)$.