14+ x - 38+ 3x • Lim V4+x - 3 X- 0

$$\begin{gathered} \underset{x\to 0}{lt}\frac{\sqrt{4+x}-\sqrt[3]{8+3x}}{x}=\underset{x\to 0}{lt}\frac{{\left(4+x\right)}^{1/2}-{\left(8+3x\right)}^{1/3}}{x} \\ =\underset{x\to 0}{lt}\frac{{\displaystyle \frac{1}{2{(4+x)}^{1/2}}}-{\displaystyle \frac{3}{3}}(8+{3x)}^{2/3}}{1}\left(L\text{'}Hospital rule\right) \\ =\underset{x\to 0}{lt}\frac{1}{2(4+4{+x)}^{1/2}}=\frac{1}{(8+{3x)}^{2/3}} \\ =\frac{1}{2\times 2}-\frac{1}{{\left(8\right)}^{2/3}} \\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\times }\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{\left(}\mathbf{8}\mathbf{\right)}}^{\mathbf{2}\mathbf{/}\mathbf{3}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{=}\mathbf{0} \end{gathered}$$