Tính giới hạn sau: \(\mathop {\lim }\limits_{x \to \infty } \frac{{\ln ({n^2} - n + 1)}}{{\ln ({n^{10}} + n + 1)}}\)

Ta có: \(\mathop {\lim }\limits_{x \to \infty } \frac{{\ln ({n^2} - n + 1)}}{{\ln ({n^{10}} + n + 1)}}\) = \(\mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left[ {{n^2}\left( {1 - \frac{1}{n} + \frac{1}{{{n^2}}}} \right)} \right]}}{{\ln \left[ {{n^{10}}\left( {1 + \frac{1}{{{n^9}}} + \frac{1}{{{n^{10}}}}} \right)} \right]}}\) = \(\mathop {\lim }\limits_{x \to \infty } \frac{{\ln {n^2} + \ln \left( {1 - \frac{1}{n} + \frac{1}{{{n^2}}}} \right)}}{{\ln {n^{10}} + \ln \left( {1 + \frac{1}{{{n^9}}} + \frac{1}{{{n^{10}}}}} \right)}}\) = \(\mathop {\lim }\limits_{x \to \infty } \frac{{2\ln n + \ln \left( {1 - \frac{1}{n} + \frac{1}{{{n^2}}}} \right)}}{{10\ln n + \ln \left( {1 + \frac{1}{{{n^9}}} + \frac{1}{{{n^{10}}}}} \right)}}\) = \(\mathop {\lim }\limits_{x \to \infty } \frac{{2 + \frac{{\ln \left( {1 - \frac{1}{n} + \frac{1}{{{n^2}}}} \right)}}{{\ln n}}}}{{10 + \frac{{\ln \left( {1 + \frac{1}{{{n^9}}} + \frac{1}{{{n^{10}}}}} \right)}}{{\ln n}}}}\) = \(\frac{{2 + 0}}{{10 + 0}} = \frac{1}{5}\) Vì \(\mathop {\lim }\limits_{n \to \infty } \frac{{\ln \left( {1 - \frac{1}{n} + \frac{1}{{{n^2}}}} \right)}}{{\ln n}} = 0\) và \(\mathop {\lim }\limits_{n \to \infty } \frac{{\ln \left( {1 + \frac{1}{{{n^9}}} + \frac{1}{{{n^{10}}}}} \right)}}{{\ln n}} = 0\).