limits - Why $lim limits_ {nto infty}left (frac{n+3}{n+4}right)^n neq 1$?

Why doesn't $\lim\limits_ {n\to \infty}\ (\frac{n+3}{n+4})^n$ equal $1$?

So this is the question.

I found it actually it equals $e^{-1}$. I could prove it, using some reordering and canceling.

However another way I took was this:

$$\lim_ {n\to \infty}\ \left(\frac{n}{n+4}+\frac{3}{n+4}\right)^n$$

with the limit of the first term going to $1$ and the second to $0$. So $(1+0)^n=1$ not $e^{-1}$.

Answer

Because $1^\infty$ is a tricky beast. Perhaps the power overwhelms the quantity that's
just bigger than $1$, but approaching $1$, and the entire expression is large. Or perhaps not...

Perhaps the power overwhelms the quantity that's just smaller than $1$, but approaching $1$, and the entire expression tends to $0$ . Or perhaps not...

In your case,

$${n+3\over n+4} = 1-{1\over n+4}.$$
And, as one can show (as you did): $$\lim\limits_{n\rightarrow\infty}(1-\textstyle{1\over n+4})^n = \lim\limits_{n\rightarrow\infty}\Bigl[ (1-\textstyle{1\over n+4})^{n+4}\cdot (1-{1\over n+4})^{-4}\Bigr] = e^{-1}\cdot1=e^{-1}.$$

Here, the convergence of $1-{1\over n+4}$ to 1 is too fast for the $n^{\rm th}$ power to drive it back down to $0$.