Question

Discuss the continuity and differentiability of f (x) = |log |x||.

Answer 1

We have,
f (x) = |log |x||

$$\begin{gathered} \left|x\right|=\left\{\begin{array}{l}\begin{array}{l}-x-\infty <x<-1\\ -x-1<x<0\end{array}\\ \begin{array}{l}x0<x<1\\ x1<x<\infty \end{array}\end{array}\right. \\ \log \left|x\right|=\left\{\begin{array}{l}\begin{array}{l}\log \left(-x\right)-\infty <x<-1\\ \log \left(-x\right)-1<x<0\end{array}\\ \begin{array}{l}\log \left(x\right)0<x<1\\ \log \left(x\right)1<x<\infty \end{array}\end{array}\right. \\ \left|\log \left|x\right|\right|=\left\{\begin{array}{l}\begin{array}{l}\log \left(-x\right)-\infty <x<-1\\ -\log \left(-x\right)-1<x<0\end{array}\\ \begin{array}{l}-\log \left(x\right)0<x<1\\ \log \left(x\right)1<x<\infty \end{array}\end{array}\right. \end{gathered}$$
$$\begin{gathered} \left(\mathrm{LHD}\mathrm{at}x=-1\right)=\underset{x\to -1^{-}}{\lim }\frac{f\left(x\right)-f\left(-1\right)}{x+1} \\ =\underset{x\to -1^{-}}{\lim }\frac{\log \left(-x\right)-0}{x+1} \\ =\underset{h\to 0}{\lim }\frac{\log \left(1+h\right)}{-1-h+1} \\ =-\underset{h\to 0}{\lim }\frac{\log \left(1+h\right)}{h}=-1 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{RHD}\mathrm{at}x=-1\right)=\underset{x\to -1^{+}}{\lim }\frac{f\left(x\right)-f\left(-1\right)}{x+1} \\ =\underset{x\to -1^{+}}{\lim }\frac{-\log \left(-x\right)-0}{x+1} \\ =\underset{h\to 0}{\lim }\frac{-\log \left(1-h\right)}{-1+h+1} \\ =\underset{h\to 0}{\lim }\frac{-\log \left(1-h\right)}{h}=1 \end{gathered}$$
Here, LHD ≠ RHD
So, function is not differentiable at x = − 1

At 0 function is not defined.
$$\begin{gathered} \left(\mathrm{LHD}\mathrm{at}x=1\right)=\underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ =\underset{x\to 1^{-}}{\lim }\frac{-\log \left(x\right)-0}{x-1} \\ =\underset{h\to 0}{\lim }\frac{-\log \left(1-h\right)}{1-h-1} \\ =-\underset{h\to 0}{\lim }\frac{\log \left(1-h\right)}{h}=-1 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{RHD}\mathrm{at}x=1\right)=\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ =\underset{x\to 1^{+}}{\lim }\frac{\log \left(x\right)-0}{x-1} \\ =\underset{h\to 0}{\lim }\frac{\log \left(1+h\right)}{1+h-1} \\ =\underset{h\to 0}{\lim }\frac{\log \left(1+h\right)}{h}=1 \end{gathered}$$
Here, LHD ≠ RHD
So, function is not differentiable at x = 1
Hence, function is not differentiable at x = 0 and ± 1
At 0 function is not defined.
So, at 0 function is not continuous.
$$\begin{gathered} \left(\mathrm{LHL}\mathrm{at}x=-1\right)=\underset{x\to -1^{-}}{\lim }f\left(x\right) \\ =\underset{x\to -1^{-}}{\lim }\log \left(-x\right) \\ =\log \left(1\right)=0 \\ \left(\mathrm{RHL}\mathrm{at}x=-1\right)=\underset{x\to -1^{+}}{\lim }f\left(x\right) \\ =\underset{x\to -1^{+}}{\lim }-\log \left(-x\right) \\ =-\log 1=0 \\ f\left(-1\right)=0 \\ \mathrm{Therefore},f\left(x\right)=\left|\log \left|x\right|\right|\mathrm{is}\mathrm{continuous}\mathrm{at}x=-1 \end{gathered}$$
$$\begin{gathered} \left(\mathrm{LHL}\mathrm{at}x=1\right)=\underset{x\to 1^{-}}{\lim }f\left(x\right) \\ =\underset{x\to 1^{-}}{\lim }-\log \left(x\right) \\ =-\log \left(1\right)=0 \\ \left(\mathrm{RHL}\mathrm{at}x=1\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right) \\ =\underset{x\to 1^{+}}{\lim }\log \left(x\right) \\ =\log 1=0 \\ f\left(1\right)=0 \\ \mathrm{Therefore},\mathrm{at}x=1,f\left(x\right)=\left|\log \left|x\right|\right|\mathrm{is}\mathrm{continuous}. \end{gathered}$$

Hence, function f (x) = |log |x|| is not continuous at x = 0