The value of $$\lim _\limits{x \rightarrow \infty}\left(\frac{x^2-2 x+1}{x^2-4 x+2}\right)^{2 x}$$ is

$$ \lim _\limits{x \rightarrow 0} \frac{a^x-b^x}{x} \text { is equal to } $$

$$ \text { The function defined by } f(x)=\left\{\begin{array}{cc} \frac{\sin x}{x}+\cos x & x>0 \\ -5 k & x=0 \\ \frac{4(1-\sqrt{1-x})}{x} & x<0 \end{array} \quad \text { is continous at } x=0, \quad \text { then } k\right. \text { equals } $$

If $$\mathop {\lim }\limits_{x \to 0} {{(1 + {a^3}) + 8{e^{1/x}}} \over {1 + (1 - {b^3}){e^{1/x}}}} = 2$$, then