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\[\lim_{x->\propto }(x+2)\times e^{\frac{x}{x-1}}-ex\]



Come risolvo questo limite? DOvrebbe venire 3e

Ecco il limite:
\[\begin{array}{l}
\mathop {\lim }\limits_{x \to \infty } \;[(x + 2)\,{e^{\frac{x}{{x - 1}}}} - e{\kern 1pt} x] = \mathop {\lim }\limits_{x \to \infty } \;[x{e^{\frac{x}{{x - 1}}}} + 2{e^{\frac{x}{{x - 1}}}} - e{\kern 1pt} x] = \mathop {\lim }\limits_{x \to \infty } \;[x{e^{\frac{x}{{x - 1}}}} - e{\kern 1pt} x + 2{e^{\frac{x}{{x - 1}}}}] = \\
= \mathop {\lim }\limits_{x \to \infty } \;[x{e^{\frac{x}{{x - 1}}}} - e{\kern 1pt} x] + \mathop {\lim \,}\limits_{x \to \infty } [2{e^{\frac{x}{{x - 1}}}}] = \mathop {\lim }\limits_{x \to \infty } \;[x({e^{\frac{x}{{x - 1}}}} - e{\kern 1pt} )] + [2{e^1}] = \mathop {\lim }\limits_{x \to \infty } \;[xe(\frac{{{e^{\frac{x}{{x - 1}}}}}}{e} - 1)] + [2e] = \\
= \mathop {\lim }\limits_{x \to \infty } \;[xe({e^{\frac{x}{{x - 1}} - 1}} - 1)] + [2e] = \mathop {\lim }\limits_{x \to \infty } \;[xe({e^{\frac{{x - x + 1}}{{x - 1}}}} - 1)] + [2e] = \mathop {\lim }\limits_{x \to \infty } \;[xe({e^{\frac{1}{{x - 1}}}} - 1)] + [2e]\; = \\
= \mathop {\lim }\limits_{x \to \infty } \;[xe(\frac{{{e^{\frac{1}{{x - 1}}}} - 1}}{{\frac{1}{{x - 1}}}})\frac{1}{{x - 1}}] + [2e] = \\
dato\;che\;:\;\frac{1}{{x - 1}} \to 0\;\;per\;\;x \to \infty \;\;allora\;\;\mathop {\lim }\limits_{x \to \infty } \frac{{{e^{\frac{1}{{x - 1}}}} - 1}}{{\frac{1}{{x - 1}}}} = 1\;\;(\lim \;notevole)\\
\mathop { = \lim }\limits_{x \to \infty } \;[xe(\frac{{{e^{\frac{1}{{x - 1}}}} - 1}}{{\frac{1}{{x - 1}}}})\frac{1}{{x - 1}}] + [2e] = \mathop {\lim }\limits_{x \to \infty } \;[\frac{{xe}}{{x - 1}}] + [2e] = \;[e] + [2e] = 3e
\end{array}\]

Saluti.

Grazie