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Un pensiero riguardo “Leandro scrive: Calcolo di limiti 1”
nella prima formula della risposta deve essere un -x nel denominatore,
o meglio il risultato è -1/6.
Il risultato finale diventa -1/12
il limite notevole è
((1+x)^n-1)/x=n
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![\[\lim_{x \to 0} \frac{\sqrt[6]{1-x}-1}{e^{2x}-1}\]](http://www.matebook.it/wp-content/ql-cache/quicklatex.com-819aaf55e4ca0ab2f20d9631c85d3e98_l3.png "Rendered by QuickLaTeX.com")
![\[\lim_{x \to 0} \frac{\sqrt[6]{1-x}-1}{x}=\frac 16\]](http://www.matebook.it/wp-content/ql-cache/quicklatex.com-f21a0fe76e69a932ffb96d2176aac32a_l3.png "Rendered by QuickLaTeX.com")
![\[\lim_{x \to 0} \frac{e^{2x}-1}{x}=2\]](http://www.matebook.it/wp-content/ql-cache/quicklatex.com-12dcccdd3cfcc54017350fb8a3979297_l3.png "Rendered by QuickLaTeX.com")
![\[\lim_{x \to 0} \frac{\sqrt[6]{1-x}-1}{e^{2x}-1}=\frac{\frac 16}{2}=\frac{1}{12}\]](http://www.matebook.it/wp-content/ql-cache/quicklatex.com-ce1e14bb7a26c7bf96b095662190b66a_l3.png "Rendered by QuickLaTeX.com")
nella prima formula della risposta deve essere un -x nel denominatore,
o meglio il risultato è -1/6.
Il risultato finale diventa -1/12
il limite notevole è
((1+x)^n-1)/x=n