Evaluate $\lim\limits_{n\rightarrow\infty}\left(\frac{n^2 + 1}{n^2 - 2}\right)^{n^2}$

Question

$$\lim\limits_{n\rightarrow\infty}\left(\frac{n^2 + 1}{n^2 - 2}\right)^{n^2}$$

Trying to solve this. At first thought it was $1$, but in wolfram it's $e^3$. Thanks

Answer 1

$$\lim_{n\to +\infty}\left(\frac{n^2+1}{n^2-2}\right)^{n^2}=\lim_{n\to +\infty}\left(\left(1+\frac{1}{\frac{n^2-2}{3}}\right)^{\frac{n^2-2}{3}}\right)^{\frac{3n^2}{n^2-2}}$$

$$=e^{\lim_{n\to +\infty}\frac{3n^2}{n^2-2}}=e^{\lim_{n\to +\infty}\left(3+\frac{6}{n^2-2}\right)}=e^3$$

Answer 2

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

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

Answer 3

Notice, we know $\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n=e$ hence,

$$\lim_{n\to \infty}\left(\frac{n^2+1}{n^2-2}\right)^{n^2}=\lim_{n\to \infty}\left(1+\frac{3}{n^2-2}\right)^{n^2}$$ Now, let $\frac{3}{n^2-2}=\frac{1}{t}\implies n^2=3t+2$ $$=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{3t+2}$$ $$=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{3t}\cdot \lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{2}$$ $$=\left(\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{t}\right)^3\cdot \lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{2}$$ $$=\left(e\right)^3\cdot \left(1\right)^{2}=\color{red}{e^3}$$