Question #34fda

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Answer 1 · 2017-04-15T16:08:40

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

enter image source here