Ciao, ti prego innanzitutto di sforzarti di usare l'equation editor per scrivere correttamente le formule matematiche. Se hai difficoltà guarda la relativa guida qui nel forum!
Riguardo al tuo limite raccogliendo i termini dominanti si ha:
\[\mathop {\lim }\limits_{x \to + \infty } \frac{{{{\ln }^2}(1 + {x^4})}}{{{{\ln }^4}(1 + {x^2})}}\mathop = \limits^{[\frac{\infty }{\infty }]} \mathop {\lim }\limits_{x \to + \infty } \frac{{{{\ln }^2}({x^4}(\frac{1}{{{x^4}}} + 1)}}{{{{\ln }^4}({x^2}(\frac{1}{{{x^2}}} + 1)}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{[\ln {{({x^4}(\frac{1}{{{x^4}}} + 1)]}^2}}}{{[\ln {{({x^2}(\frac{1}{{{x^2}}} + 1)]}^4}}} = \]
\[ = \mathop {\lim }\limits_{x \to + \infty } \frac{{{{[\ln ({x^4}) + \ln (\frac{1}{{{x^4}}} + 1)]}^2}}}{{{{[\ln ({x^2}) + \ln (\frac{1}{{{x^2}}} + 1)]}^4}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{{{[4\ln (x) + \ln (\frac{1}{{{x^4}}} + 1)]}^2}}}{{{{[2\ln (x) + \ln (\frac{1}{{{x^2}}} + 1)]}^4}}} = \]
\[ = \mathop {\lim }\limits_{x \to + \infty } \frac{{{{[\ln (x)\,(4 + \frac{{\ln (\frac{1}{{{x^4}}} + 1)}}{{\ln (x)}})]}^2}}}{{{{[\ln (x)\,(2 + \frac{{\ln (\frac{1}{{{x^2}}} + 1)}}{{\ln (x)}})]}^4}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{{{\ln }^2}(x)\,{{(4 + \frac{{\ln (\frac{1}{{{x^4}}} + 1)}}{{\ln (x)}})}^2}}}{{{{\ln }^4}(x)\,{{(2 + \frac{{\ln (\frac{1}{{{x^2}}} + 1)}}{{\ln (x)}})}^4}}} = \]
\[ = \mathop {\lim }\limits_{x \to + \infty } \frac{{{{(4 + \frac{{\ln (\frac{1}{{{x^4}}} + 1)}}{{\ln (x)}})}^2}}}{{{{\ln }^2}(x)\,{{(2 + \frac{{\ln (\frac{1}{{{x^2}}} + 1)}}{{\ln (x)}})}^4}}} = \frac{{{{(4 + \frac{{\ln (0 + 1)}}{{ + \infty }})}^2}}}{{( + \infty ){{(2 + \frac{{\ln (0 + 1)}}{{ + \infty }})}^4}}} = \frac{{{{(4 + \frac{{\ln (1)}}{{ + \infty }})}^2}}}{{( + \infty ){{(2 + \frac{{\ln (1)}}{{ + \infty }})}^4}}} = \]
\[ = \frac{{{{(4 + \frac{0}{{ + \infty }})}^2}}}{{( + \infty ){{(2 + \frac{0}{{ + \infty }})}^4}}} = \frac{{{{(4 + 0)}^2}}}{{( + \infty ){{(2 + 0)}^4}}} = \frac{{{{(4)}^2}}}{{( + \infty ){{(2)}^4}}} = \frac{{16}}{{ + \infty }} = 0\]
Saluti.
