Special limit: $1^\infty$

Question

This is my limit,

$\lim\limits_{n \to \infty } (1+\frac{1}{n^2})^{3n^2+4}$.

If I place the infinity instead of $n$, I get $1^\infty$.

I know that this is a special limit, but how I need to continue?

Answer 1

We use the limit $\lim\limits_{n \to \infty } (1+\frac{1}{n})^{n}= e$, it does not change if put $n^2$. $$ \lim\limits_{n \to \infty } (1+\frac{1}{n^2})^{3n^2+4} = \lim\limits_{n \to \infty } (1+\frac{1}{n^2})^{3n^2}\cdot(1+\frac{1}{n^2})^{4} = \\ \lim\limits_{n \to \infty } (1+\frac{1}{n^2})^{3n^2}\cdot\lim\limits_{n \to \infty }(1+\frac{1}{n^2})^{4} = \\ \left(\lim\limits_{n \to \infty } (1+\frac{1}{n^2})^{n^2}\right)^3\cdot\lim\limits_{n \to \infty }(1+\frac{1}{n^2})^{4} = e^3\cdot 1 $$

Answer 2

Hint: since $a_n{}^{b_n}=e^{b_n\ln a_n}$ then we get $$ \lim a_n{}^{b_n}=e^{\lim b_n\ln a_n}. $$

Answer 3

Notice, let $n^2=t$ $$\lim_{n\to \infty}\left(1+\frac{1}{n^2}\right)^{3n^2+4}=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{3t+4}$$ $$=\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{3t}\cdot \lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{4}$$ $$=\left(\lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{t}\right)^3\cdot \lim_{t\to \infty}\left(1+\frac{1}{t}\right)^{4}$$ $$=\left(e\right)^3\cdot \left(1\right)^{4}=\color{red}{e^{3}}$$

Answer 4

Hint

Another way, considering $$A=(1+\frac{1}{n^2})^{3n^2+4}$$ Take logarithms $$\log(A)=(3n^2+4)\log(1+\frac{1}{n^2})$$ Now, remember that if $x$ is small $\log(1+x)\approx x$. Replace $x$ by $\frac{1}{n^2}$.

I am sure that you can take from here.