Why doesn't lim 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.
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