Question

Question #34fda

Answer 1

Because `lim_(x->0^-)f(x)!=lim_(x->0^+)f(x)` and is different from `f(0)=0`.

We have that

`lim_(x->0^-)[absx]/(x^2+2x)=lim_(x->0^-)[-x]/[x^2+2x]=lim_(x->0^-)-[1]/[x+2]=-1/2`

and

`lim_(x->0^+)[absx]/(x^2+2x)=lim_(x->0^+)[x]/[x^2+2x]=lim_(x->0^+)[1]/[x+2]=1/2`

The graph of f(x) around zero is