Evaluate `lim_(x->0^+)(x^2+x)|__ 1/x __|` :

Note that for 0<x<1 we have `(x^2+x) |__ 1/x __| <=(x^2+x)1/x`

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

Also `(x^2+x) |__ 1/x __|>=(x^2+x)(1/x-1)`

And `lim_(x->0^+)(x^2+x)(1/x-1)=lim_(x->0^+)1-x^2=1`

Thus by the squeeze theorem, `lim_(x->0^+)(x^2+x) |__ 1/x __| =1`

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