Evaluate $\lim\limits_{n\rightarrow\infty}\left(\frac{n^2 + 1}{n^2 - 2}\right)^{n^2}$ $$\lim\limits_{n\rightarrow\infty}\left(\frac{n^2 + 1}{n^2 - 2}\right)^{n^2}$$
Trying to solve this. At first thought it was $1$, but in wolfram it's $e^3$. Thanks

Since $$\lim _{ n\rightarrow \infty  }{ { \left( 1+\frac { 1 }{ n }  \right)  }^{ n } } =e$$
so
$$\lim _{ n\rightarrow \infty  }{ { \left( \frac { { n }^{ 2 }+1 }{ { n }^{ 2 }-2 }  \right)  }^{ { n }^{ 2 } } } =\lim _{ n\rightarrow \infty  }{ { \left( \frac { { n }^{ 2 }-2+3 }{ { n }^{ 2 }-2 }  \right)  }^{ { n }^{ 2 } } } =\lim _{ n\rightarrow \infty  }{ { \left( 1+\frac { 3 }{ { n }^{ 2 }-2 }  \right)  }^{ { n }^{ 2 } } } =\\ =\lim _{ n\rightarrow \infty  }{ { \left( 1+\frac { 1 }{ \frac { { n }^{ 2 }-2 }{ 3 }  }  \right)  }^{ \frac { { n }^{ 2 }-2 }{ 3 } \frac { { 3n }^{ 2 } }{ { n }^{ 2 }-2 }  } } ={ e }^{ \lim _{ n\rightarrow \infty  }{ \frac { 3{ n }^{ 2 } }{ { n }^{ 2 }-2 }  }  }={ e }^{ \lim _{ n\rightarrow \infty  }{ \frac { 3{ n }^{ 2 } }{ { n }^{ 2 }\left( 1-\frac { 2 }{ { n }^{ 2 } }  \right)  }  }  }={ e }^{ 3 }$$
