Match the following $\underset{x\to 0}{lim} f\left(x\right)$ where ‘f’ is given in column I with corresponding limit values in
column-II.

$$\begin{gathered} \left(A\right) \\ =\underset{x\to 0}{Lim}\frac{\sin 3x}{\log \left(1+5x\right)} \\ =\underset{x\to 0}{Lim}\frac{\sin 3x}{3x}.3x \underset{x\to 0}{Lim}\frac{5x}{l\mathrm{og}\left(1+5x\right)}\frac{1}{5x} \\ =\frac{3}{5} \\ \left(B\right) \\ =\underset{x\to 0}{Lim}\frac{e^{\sin 3x}-1}{\log \left(1+\tan 2x\right)} \\ =\underset{x\to 0}{Lim}\frac{e^{\sin 3x}-1}{\sin 3x}.\sin 3x \underset{x\to 0}{Lim}\frac{\tan 2x}{\log \left(1+\tan 2x\right)}.\frac{1}{\tan 2x} \\ =\frac{3}{2} \\ \left(C\right) \\ =\underset{x\to 0}{Lim}\frac{\log \left(2-\cos 2x\right)}{{\left(\log \left(\sin 3x+1\right)\right)}^2} \\ =\underset{x\to 0}{Lim}\frac{\log \left(1+2{\sin }^{2}x\right)}{2{\sin }^{2}x}.2{\sin }^{2}x \underset{x\to 0}{Lim}\frac{{\sin }^{2}3x}{{\left(\log \left(\sin 3x+1\right)\right)}^2}.\frac{1}{{\sin }^{2}3x} \\ =\frac{2}{9} \end{gathered}$$