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![\[\lim_{x \to 0^+}\left(\frac x2 \right)^{\frac{-3}{ln(x)}}\]](http://www.matebook.it/wp-content/ql-cache/quicklatex.com-5104395b1be980ac47a0818932fbcb7d_l3.png "Rendered by QuickLaTeX.com")
![\[\lim_{x \to 0^+}e^{\frac{-3}{ln(x)} \cdot ln \frac x2}=\]](http://www.matebook.it/wp-content/ql-cache/quicklatex.com-ac608fdf08d1e6c9b0b1ea55d2da1372_l3.png "Rendered by QuickLaTeX.com")
![\[\lim_{x \to 0^+}e^{\frac{-3(ln(x)-ln2)}{ln(x)} }\simeq\]](http://www.matebook.it/wp-content/ql-cache/quicklatex.com-cdd3d0fe7bbac060696a62b3f5bb2df6_l3.png "Rendered by QuickLaTeX.com")
![\[\lim_{x \to 0^+}e^{\frac{-3ln(x)}{ln(x)} }=e^{-3}\]](http://www.matebook.it/wp-content/ql-cache/quicklatex.com-351ffefcc7848cb239a02a522af029b1_l3.png "Rendered by QuickLaTeX.com")
Vorrei sapere perchè nell’ultimo limite si annulla – ln2.
Grazie
Perchè il log 2 è un numero finito mentre il log (0+) tende a meno infinito.