$\begin{align}
& \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,(2f(x)+1)=5\Rightarrow \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,2f(x)=4\Rightarrow \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=2 \\
& ba\,tavajoh\,b\,ne\bmod ar\Rightarrow \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,g(x)=\frac{-1}{2} \\
& \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\frac{\sqrt{({{f}^{3}}-2g)(x)}}{(f.g)(x)+3}=\frac{\sqrt{{{(\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x))}^{3}}-2(\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,g(x))}}{\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x).\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,g(x)+3} \\
& =\frac{\sqrt{{{(2)}^{3}}-2(\frac{-1}{2})}}{2(\frac{-1}{2})+3}=\frac{\sqrt{8+1}}{-1+3}=\frac{3}{2}=1/5 \\
\end{align}$