Show that the function `f(x)={x^msin(1/x),\ \ \ x!=0 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0` is neither continuous nor differentiable, if `mlt=0`

Explanation

Given function
`f(x)={{:(x^msin(1/x), xne0),(0,x=0):}`
`LHL=lim_(x->0^-)f(x)`
`=lim_(h->0)f(0-h)`
`=lim_(h->0)(-h)^msin(-1/h)`
which is tends to infinity
`RHL=lim_(x->0^+)f(x)`
`=lim_(h->0)f(0+h)`
`=lim_(h->0)(h)^msin(1/h)`
which is tends to infinity

Since, f(x) is not continuous at` x = 0`
For Differentiability at `x = 0 `
`LHD` at `x=0`
`lim_(x->0^-)(f(x)-f(0))/(x-0)`
`=lim_(h->0)(f(0-h)-f(0))/(0-h-0)`
`=>lim_(h->0)((-h)^msin(-1/h))/-h`
`=>lim_(h->0)((-h)^(m-1)sin(-1/h))`
`=`not defined
`RHD` at `x=0`
`lim_(x->0^+)(f(x)-f(0))/(x-0)`
`=lim_(h->0)(f(0+h)-f(0))/(0+h-0)`
`=>lim_(h->0)((h)^msin(1/h))/h`
`=>lim_(h->0)((h)^(m-1)sin(1/h))`
`=0`
clearly,`LHDneRHD`
hence, f(x) is continuous but not differentiable at `x=0`
, if `mlt=0`