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

Question:

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

Solution:

We have,

f (x) = |log |x||

$|x|=\left\{\begin{array}{cl}-x & -\infty$\log |x|=\left\{\begin{array}{cc}\log (-x) & -\infty

$|\log | x||=\left\{\begin{array}{lc}\log (-x) & -\infty$(\mathrm{LHD}$at$x=-1)=\lim _{x \rightarrow-1^{-}} \frac{f(x)-f(-1)}{x+1}=\lim _{x \rightarrow-1^{-}} \frac{\log (-x)-0}{x+1}=\lim _{h \rightarrow 0} \frac{\log (1+h)}{-1-h+1}=-\lim _{h \rightarrow 0} \frac{\log (1+h)}{h}=-1(\mathrm{RHD}$at$x=-1)=\lim _{x \rightarrow-1^{+}} \frac{f(x)-f(-1)}{x+1}=\lim _{x \rightarrow-1^{+}} \frac{-\log (-x)-0}{x+1}=\lim _{h \rightarrow 0} \frac{-\log (1-h)}{-1+h+1}=\lim _{h \rightarrow 0} \frac{-\log (1-h)}{h}=1$Here, LHD ≠ RHD So, function is not differentiable at x = − 1 At 0 function is not defined.$(\mathrm{LHD}$at$x=1)=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{-}} \frac{-\log (x)-0}{x-1}=\lim _{h \rightarrow 0} \frac{-\log (1-h)}{1-h-1}=-\lim _{h \rightarrow 0} \frac{\log (1-h)}{h}=-1(\mathrm{RHD}$at$x=1)=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1^{+}} \frac{\log (x)-0}{x-1}=\lim _{h \rightarrow 0} \frac{\log (1+h)}{1+h-1}=\lim _{h \rightarrow 0} \frac{\log (1+h)}{h}=1$Here, LHD ≠ RHD So, function is not differentiable at x = 1 Hence, function is not differentiable at x = 0 and ± 1 At 0 function is not defined. So, at 0 function is not continuous.$(\mathrm{LHL}$at$x=-1)=\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{-}} \log (-x)=\log (1)=0(\mathrm{RHL}$at$x=-1)=\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}}-\log (-x)=-\log 1=0f(-1)=0$Therefore,$f(x)=|\log | x||$is continuous at$x=-1$'$(\mathrm{LHL}$at$x=1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}-\log (x)=-\log (1)=0(\mathrm{RHL}$at$x=1)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} \log (x)=\log 1=0f(1)=0$Therefore, at$x=1, f(x)=|\log | x||$is continuous. Hence, function$f(x)=\| \log |x| \mid$is not continuous at$x=0\$