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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