Why does $\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n=e$ but $\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n=e^{-1}$? Why does $$\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n=e$$ but $$\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n=e^{-1}$$
Shouldn't the limits be the same since $\left(1+\frac{1}{n}\right) \to 1$?

Expand both using the binomial theorem
$$\left(1+\frac 1n\right)^n=1^n+n\cdot\frac 1n\cdot1^{n-1}+\binom n2\left(\frac 1n\right)^21^{n-2}+\binom n3\left(\frac 1n\right)^31^{n-3}+\dots =$$$$=1+1+\frac {n(n-1)}{2n^2}+\frac {n(n-1)(n-2)}{6n^3}+\dots$$while $$\left(1-\frac 1n\right)^n=1-1+\frac {n(n-1)}{2n^2}-\frac {n(n-1)(n-2)}{6n^3}+\dots$$
Subtract the second from the first and you get $$2+\frac {n(n-1)(n-2)}{3n^3}+\dots$$ a sum of positive terms. So the difference between the two expressions is at least $2$. If the limits exist (they do), the difference between the limits must be at least $2$.
If you analyse the difference you will note that it is increasing with $n$ (but bounded) so even though the $(1\pm \frac 1n)$ part tends to zero, raising to the $n^{th}$ power in this case gives expressions which get further apart rather than closer together.
