Why does $\lim_{n\to \infty} (1+\frac{1}{n!})^{2n} = \lim_{n\to \infty}\big( (1+\frac{1}{n!})^{n!}\big)^{\frac2{(n-1)!}}=e^0$?

Question

I understand the algebraic manipulation and I'm assuming that the thinking is: $\lim_{n\to \infty}\left( \left(1+\frac{1}{n!}\right)^{n!}\right) = e$ and $\lim_{n \to \infty} \frac2{(n-1)!} = 0$.

However, I don't understand why we can break the expression $\lim_{n\to \infty}\left( \left(1+\frac{1}{n!}\right)^{n!}\right)^{\frac2{(n-1)!}}$, into $\lim_{n\to \infty}\left( \left(1+\frac{1}{n!}\right)^{n!}\right)$ and $\lim_{n \to \infty} \frac2{(n-1)!}$.

It seems similar to breaking $\lim_{n \to \infty} \big(1+\frac{1}{n}\big)^n$ into $\lim_{n \to \infty} \big(1+\frac{1}{n}\big)$ and $\lim_{n \to \infty} (n)$ and saying that $\lim_{n \to \infty} \big(1+\frac{1}{n}\big)^n = 1$, which is obviously wrong.

So why are we in this case allowed to consider separately $\lim_{n\to \infty}\left( \left(1+\frac{1}{n!}\right)^{n!}\right)$ and $\lim_{n \to \infty} \frac2{(n-1)!}$?

EDIT: I understand that if the exponent were a constant it would make sense to separate the two limits. What confuses me in this case is that the exponent also depends on $n$, so why are we allowed to separate the two?

Answer 1

Consider $\lim f(x)^{g(x)}$. We rewrite as $$\lim e^{(\ln f(x))g(x)}=e^{\lim g(x)\ln f(x)}$$ where we may pull the limit into the exponential because it is a continuous function. We turn our attention to the exponent. Using the standard limit theorem, we may write $$\lim g(x)\ln f(x)=(\lim g(x))(\lim \ln f(x))$$ provided both limits exist, and the result is not indeterminate. In this case, $\lim g(x)=0$ and $\lim \ln f(x)=1$, so $$\lim g(x)\ln f(x)=0\cdot 1=0$$

Answer 2

We can show the validity of the limit in a straightforward way. We recall that for $0<x<1$, the sequence of functions $e_n(x)$ given by

$$e_n(x)=\left(1+\frac xn\right)^{2n}$$

is monotonically increasing for $0<x<1$ and satisfies the inequalities

$$1\le e_n(x)\le e^{2x}$$

If we let $x=n/n!$, then we find that

$$1\le \left(1+\frac{1}{n!}\right)^{2n}\le e^{2n/n!}$$

And now use of the Squeeze Theorem gives the expected result.

Answer 3

I assume you are familiar with the fact that if $(a_n)_{n\in N}$ converges to $A$ and $(b_n)_{n\in N}$ converges to $B$ then $(a_nb_n)_{n\in N}$ converges to $A B$. Therefore,if $\lim_{n\to \infty}x_n=X>0$ and $\lim_{n\to \infty}y_n=Y$ then $$\lim_{n\to \infty}x_n^{y_n}=X^Y.$$ PROOF: Let $z_n=y_n\log x_n .$ The logarithm function is continuous, so $\lim_{n\to \infty}(\log x_n)=\log X$. Therefore $$Y\log X=\lim_{n\to \infty}(y_n\log x_n)=\lim_{n\to \infty}z_n.$$ The exponential function is continuous.Therefore $$X^Y=\exp (Y\log X)=\lim_{n\to \infty} \exp (z_n) =\lim_{n\to \infty}x_n^{y_n}\;.$$ In your Q, let $x_n=(1+1/n!)^{n!}$ and $ y_n=2/(n-1)!$.