How do I begin to evaluate this limit: $$\lim_{n\to \infty} \left(1-\frac{1}{n^2-4}\right)^{3n^2+5}$$
Does it equal to $e^{-1}$? (Please don't use ln.)
Thanks a lot.
How do I begin to evaluate this limit: $$\lim_{n\to \infty} \left(1-\frac{1}{n^2-4}\right)^{3n^2+5}$$
Does it equal to $e^{-1}$? (Please don't use ln.)
Thanks a lot.
Let $m=n^2-4$, and
$$ \lim_{m\rightarrow\infty}\left(1-\frac1m\right)^{3m+17}=\left[\lim_{m\rightarrow\infty}\left(1-\frac1m\right)^{m}\right]^3=\frac1{e^3}$$
You can also use the well known series expansion of $\exp(-x) = 1 - 1/x + O(x^{-2})$:
$$\begin{align} \lim_{n\to \infty} \left(1-\frac{1}{n^2-4}\right)^{3n^2+5} &= \lim_{n\to \infty} \exp\left(-\frac{3n^2+5}{n^2-4}\right) + O(n^{-2})\\ &= \lim_{n\to \infty} \exp\left(-3 + O(n^{-2})\right)\\ &= \exp(-3) \end{align}$$
Let $v\longrightarrow \infty$ and $u\longrightarrow 1$ , then $\lim _{n\rightarrow \infty}u^v\longrightarrow \exp (\lim_{n\rightarrow \infty} v(u-1))$.
So, we have $$\lim_{n\to \infty} \left(1-\frac{1}{n^2-4}\right)^{3n^2+5}=exp(\lim_{n\to \infty} \left(1-\frac{1}{n^2-4}-1\right)({3n^2+5})=exp(\lim_{n\to \infty} \left(-\frac{3n^2+5}{n^2-4}\right))\\ =exp(-3)$$