Show that the function `f(x)={x^m sin(1/x) , 0 ,x != 0,x=0` is differentiable at `x=0`,if `m > 1`

Explanation

For differentiablility at `x=0`

`(L H D text{ at } x=0)=lim _{x -> 0^{-}} frac{f(x)-f(0)}{x-0}`

`=lim _{h -> 0} frac{f(0-h)-f(0)}{0-h}`

`=lim _{h -> 0} {(0-h)^m sin(1/(0-h))-0}/{0-h}`

`=lim _{h -> 0} (-h)^{m-1}sin{-1/h}`

If `m>1` ,

`(L H D text{ at } x=0)=0 `

And

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