$$\lim _\limits{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}=$$

If $$f(x)$$ is continuous on its domain $$[-2,2]$$, where

$$f(x)=\left\{\begin{array}{cc} \frac{\sin a x}{x}+3 & , \text { for }-2 \leq x<0 \\ 2 x+7 & , \text { for } 0 \leq x \leq 1 \\ \sqrt{x^2+8}-b & , \text { for } 1< x \leq 2 \end{array}\right.$$ $$\text { then the value of } 2 a+3 b \text { is }$$

The function $$\mathrm{f}(\mathrm{t})=\frac{1}{\mathrm{t}^2+\mathrm{t}-2}$$ where $$\mathrm{t}=\frac{1}{x-1}$$ is discontinuous at

If $$\mathrm{f}(x)=3 x^{10}-7 x^8+5 x^6-21 x^3+3 x^2-7$$, then $$\lim _\limits{\alpha \rightarrow 0} \frac{f(1-\alpha)-f(1)}{\alpha^3+3 \alpha}=$$