Question

Show that $\underset{x\to 2^{-}}{\lim }\frac{x}{\left[x\right]}\ne \underset{x\to 2^{+}}{\lim }\frac{x}{\left[x\right]}.$

Answer 1

$$\begin{gathered} \underset{x\to 2^{-}}{\lim }\frac{x}{\left[x\right]} \\ \mathrm{Let}x=2-h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\frac{\left(2-h\right)}{\left[2-h\right]} \\ =\frac{2}{1} \\ \underset{x\to 2^{+}}{\lim }\frac{x}{\left[x\right]} \\ \mathrm{Let}x=2+h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\frac{\left(2+h\right)}{\left[2+h\right]} \\ =\frac{2}{2} \\ =1 \\ \therefore \underset{x\to 2^{-}}{\lim }\frac{x}{\left[x\right]}\ne \underset{x\to 2^{+}}{\lim }\frac{x}{\left[x\right]} \end{gathered}$$