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

Let $f(x)$ be a differentiable function such that $f(0)=0$ and $f^{\prime}(0)=20$. For $x \in \left(0, \frac{\pi}{2}\right]$,if $A(x)=2 f(x) \operatorname{cosec} 4 x+4 f(x)\left(\cos ^2 x+1\right)-4 \cos ^2 x$,then $\lim _{x \rightarrow 0} A(x)=$

$\mathop {\lim }\limits_{x \to a} f(x) \cdot g(x)$ exists,if

$\mathop {\lim }\limits_{x \to \infty } \frac{{{{\cot }^{ - 1}}\left( {\sqrt {x + 1} - \sqrt x } \right)}}{{{{\sec }^{ - 1}}\left\{ {{{\left( {\frac{{2x + 1}}{{x - 1}}} \right)}^x}} \right\}}}$ is equal to-

$\lim _{x \rightarrow 2}\left(\sum_{n=1}^{9} \frac{x}{n(n+1) x^{2}+2(2 n+1) x+4}\right)$ is equal to :

If $a, b, c$ and $d$ are positive,then $\lim_{x \to \infty} \left(1 + \frac{1}{a+bx}\right)^{c+dx}$ is equal to: