Question

Showthat:limx→±∞(1+1/x)=e

Answer 1

You mean \(\lim_{x\rightarrow \infty} (1+{1 \over x})^x=e\)?

Answer 2

Ah, this one's huge. Not hard, but lot of calcs to do. The proof (the one I studied at university at least) is divided in three parts:

1) Prove that An (sequence) given by lim x-> + infinity (1+1/n)^n is increasing
2) Prove that An is limited by proving that there is a sequence Bn that is decreasing and is > An.
3) Prove that The increasing sequence An and the decreasing sequence Bn are going to the same limit "e".

You can find the complete proof around the web of course, but this is the basic idea (At least this is the proof I studied, there might be others! ).

Answer 3

He doesn't need to do all that.
\[\lim_{x \rightarrow \pm \infty}(1+{1 \over x})^x=e^{\lim_{x \rightarrow \pm \infty} x \log (1+{1 \over x})}\]
You can show by L'hopital's rule that \(\lim_{x\to \pm \infty} x\log(1+1/x)=1\), which implies that \[\lim_{x \rightarrow \pm \infty}(1+{1 \over x})^x=e.\]

Answer 4

Well, of course. But I was talking about the proof you do with sequences, then connect to the limits afterwards. You study de L'hopital with derivatives which usually comes afterwards.