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![\[\lim_{x \to 0^-} \left(\frac {1}{x^2-2x}+\frac {1}{x^2+x} \right)=\]](https://www.matebook.it/wp-content/ql-cache/quicklatex.com-c85d238cba4bbd13bb4f613a2d843321_l3.png "Rendered by QuickLaTeX.com")
![\[=\lim_{x \to 0^-} \left(\frac {1}{x(x-2)}+\frac {1}{x(x+1)} \right)=\]](https://www.matebook.it/wp-content/ql-cache/quicklatex.com-369d1e6764aab606a43748dc80a6b138_l3.png "Rendered by QuickLaTeX.com")
![\[=\lim_{x \to 0^-} \left(\frac {x+1+x-2}{x(x-2)(x+1)} \right)=\]](https://www.matebook.it/wp-content/ql-cache/quicklatex.com-9b2e12474d6dc8810308bac5e46cf895_l3.png "Rendered by QuickLaTeX.com")
![\[=\lim_{x \to 0^-} \left(\frac {2x-1}{x(x-2)(x+1)} \right)=\frac{-1}{0^+}=-\infty\]](https://www.matebook.it/wp-content/ql-cache/quicklatex.com-6620e1f47262d7265960772a4112dc2a_l3.png "Rendered by QuickLaTeX.com")
Potrebbe spiegarmi come viene 0+ al denominatore nell’ultimo passaggio.
Grazie
Il prodotto dei tre termini al denominatore fa 0, di cui, due sono negativi, e uno positivo.