Question

Find:

$$\begin{gathered} \left(\mathrm{i}\right)\underset{x\to 2}{\lim }\left[x\right] \\ \left(\mathrm{ii}\right)\underset{x\to \frac{5}{2}}{\lim }\left[x\right] \\ \left(\mathrm{iii}\right)\underset{x\to 1}{\lim }\left[x\right] \end{gathered}$$

Answer 1

(i)

$$\begin{gathered} \underset{x\to 2}{\lim }\left[x\right] \\ \mathrm{LHL}: \\ \underset{x\to 2^{-}}{\lim }\left[x\right] \\ \mathrm{Let}x=2-h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left[2-h\right] \\ =1 \\ \mathrm{RHL}: \\ \underset{x\to 2^{+}}{\lim }\left[x\right] \\ \mathrm{Let}x=2+h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left[2+h\right] \\ =2 \\ \mathrm{LHL}\ne \mathrm{RHL} \\ \mathrm{Thus},\underset{x\to 2}{\lim }\left[x\right]\mathrm{does}\mathrm{not}\mathrm{exist}. \end{gathered}$$


(ii)
$$\begin{gathered} \underset{x\to \frac{5}{2}}{\lim }\left[x\right] \\ \mathrm{LHL} \\ \underset{x\to {\frac{5}{2}}^{-}}{\lim }\left[x\right] \\ \mathrm{Let}x=\frac{5}{2}-h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left[\frac{5}{2}-h\right] \\ =2 \\ \mathrm{RHL}: \\ \underset{x\to {\frac{5}{2}}^{+}}{\lim }\left[x\right] \\ \mathrm{Let}x=\frac{5}{2}+h,\mathrm{where}h\to 0. \\ \Rightarrow \underset{h\to 0}{\lim }\left[\frac{5}{2}+h\right] \\ \therefore \underset{x\to \frac{5}{2}}{\lim }\left[x\right]=2 \\ =2 \\ \therefore \underset{x\to \frac{5}{2}}{\lim }\left[x\right]=2 \end{gathered}$$

(iii)
$$\begin{gathered} \underset{x\to 1}{\lim }\left[x\right] \\ \mathrm{LHL}: \\ \underset{x\to 1^{-}}{\lim }\left[x\right] \\ \mathrm{Let}x=1-h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left[1-h\right] \\ =0 \\ \mathrm{RHL}: \\ \underset{x\to 1^{+}}{\lim }\left[x\right] \\ \mathrm{Let}x=1+h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left[1+h\right]=1 \\ \mathrm{LHL}\ne \mathrm{RHL} \\ \mathrm{Thus},\underset{x\to 1}{\lim }\left[x\right]\mathrm{does}\mathrm{not}\mathrm{exist}. \end{gathered}$$