If $f(1) = 1$ and $f'(1) = 4,$ then the value of $\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$ is

Evaluate the limit: $\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)^{\frac{x}{x+1-e^x}}$

If $f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$,then as $x$ approaches $a$,the limit of $\frac{g(x) f(a)-g(a) f(x)}{x-a}$ is

$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{{{\sin }^{ - 1}}x}} = $

$\mathop {\lim }\limits_{x \to 0} x \log (\sin x) = $

If $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \infty$ and $\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k$,then $12 k=$