If $f(x) = \frac{x(a^x - 1)}{1 - \cos x}$ and $g(x) = \frac{x(1 - a^x)}{a^x(\sqrt{1 - x^2} - \sqrt{1 + x^2})}$,then $\lim_{x \to 0} (f(x) - g(x)) = $

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 $\lim _{x \rightarrow 0} A(x)=$

$\lim _{x \rightarrow \infty} \frac{e^{x^4}-1}{e^{x^4}+1} = $

$\lim _{n \rightarrow \infty}\left\{n-\sqrt{n^2-4 n}\right\}=$

The value of $\lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$ is equal to

If $\mathop {Lim}\limits_{x \to 0} \frac{\ln(3 + x) - \ln(3 - x)}{x} = k$,the value of $k$ is