If $\mathrm{L}=\underset{\mathrm{x}\to 2}{lim} \frac{\sqrt[3]{60+{\mathrm{x}}^2}-4}{\sin (\mathrm{x}-2)}$, then the value of 1/L is ______ .

We have 
$\begin{array}{l}\underset{\mathrm{x}\to 2}{lim} \frac{\sqrt[3]{60+{\mathrm{x}}^2}-\sqrt[3]{64}}{\sin (\mathrm{x}-2)}\\ =\underset{\mathrm{x}\to 2}{lim} \frac{\left[{\left(60+{\mathrm{x}}^2\right)}^{\frac{1}{3}}-{64}^{\frac{1}{3}}\right]\left[{\left(60+{\mathrm{x}}^2\right)}^{\frac{2}{3}}+{\left(60+{\mathrm{x}}^2\right)}^{\frac{1}{3}}{64}^{\frac{1}{3}}+{64}^{\frac{2}{3}}\right]}{(\mathrm{x}-2)\frac{\sin (\mathrm{x}-2)}{(\mathrm{x}-2)}\left[{\left(60+{\mathrm{x}}^2\right)}^{\frac{2}{3}}+{\left(60+{\mathrm{x}}^2\right)}^{\frac{1}{3}}{64}^{\frac{1}{3}}+{64}^{\frac{2}{3}}\right]}\\ =\underset{\mathrm{x}\to 2}{lim} \frac{60+{\mathrm{x}}^2-64}{(\mathrm{x}-2)[16+4\times 4+16]}\\ =\underset{\mathrm{x}\to 2}{lim} \frac{(\mathrm{x}-2)(\mathrm{x}+2)}{48(\mathrm{x}-2)}\\ =\underset{\mathrm{x}\to 2}{lim} \frac{\mathrm{x}+2}{48}\\ =\frac{4}{48}=\frac{1}{12}\end{array}$