Ciao,
\[\mathop {\lim }\limits_{x \to + \infty } \ln (1 + \frac{1}{{{x^2}}})\frac{{x\,{2^{2x}} - sen(x) + {x^4}}}{{{4^x} - arctg(x)}} = [\frac{\infty }{\infty }]\]
per questo limite è sufficiente raccogliere i termini dominanti tendenti ad infinito per eliminare la forma indeterminata previo un disaccoppiamento della prima parte del limite con un limite notevole. Ecco tutti i passaggi:
\[\mathop {\lim }\limits_{x \to + \infty } \ln (1 + \frac{1}{{{x^2}}})\frac{{x\,{2^{2x}} - sen(x) + {x^4}}}{{{4^x} - arctg(x)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\ln (1 + \frac{1}{{{x^2}}})}}{{\frac{1}{{{x^2}}}}}\frac{1}{{{x^2}}}\frac{{x\,{2^{2x}} - sen(x) + {x^4}}}{{{4^x} - arctg(x)}} = \]
\[ = \mathop {\lim }\limits_{x \to + \infty } \frac{{\ln (1 + \frac{1}{{{x^2}}})}}{{\frac{1}{{{x^2}}}}} \cdot \mathop {\lim }\limits_{x \to + \infty } \frac{{x\,{2^{2x}} - sen(x) + {x^4}}}{{{x^2}[{4^x} - arctg(x)]}} = 1 \cdot \mathop {\lim }\limits_{x \to + \infty } \frac{{x\,{2^{2x}}[1 - \frac{{sen(x)}}{{x\,{2^{2x}}}} + \frac{{{x^4}}}{{x\,{2^{2x}}}}]}}{{{x^2}{4^x}[1 - \frac{{arctg(x)}}{{{4^x}}}]}} = \]
\[ = \mathop {\lim }\limits_{x \to + \infty } \frac{{{2^{2x}}[1 - \frac{{sen(x)}}{{x\,{2^{2x}}}} + \frac{{{x^3}}}{{{2^{2x}}}}]}}{{x\,{2^{2x}}[1 - \frac{{arctg(x)}}{{{4^x}}}]}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{1 - \frac{{sen(x)}}{{x\,{2^{2x}}}} + \frac{{{x^3}}}{{{2^{2x}}}}}}{{x\,[1 - \frac{{arctg(x)}}{{{4^x}}}]}} = \frac{{1 - 0 + 0}}{{ + \infty \,[1 - 0]}} = \frac{1}{{ + \infty }} = 0\]
Saluti.
