Why does $\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n=e$ but $\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n=e^{-1}$?

Question

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 $\left(1+\frac{1}{n}\right) \to 1$?

Answer 1

If the limits were the same, they would have to be $1$, since $$\left(1+\frac1n\right)^n>1$$ for all $n$, while $$\left(1-\frac1n\right)^n<1$$ for all $n$.

So if you expect the first limit to be $e$ and you know that $e>1$, you cannot expect the second one to be equal to it.

Edit: (to incorporate two of Paramanad Singh comments). One can obtain from Bernoulli's Inequality that $$\left(1+\frac1n\right)^n\geq2,\ \ n\geq1.$$ Which implies that $e\geq2$ as soon as one knows that the limits exists. This already shows that the limits cannot be the same. More interestingly, from $$ \left(1-\frac{1}{n}\right)^{n}=\dfrac{1}{\left(1+\dfrac{1}{n-1}\right)^{n-1}}\cdot\left(1-\frac{1} {n}\right),n>1 $$ one gets that $$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{n}=e^{-1}.$$

Answer 2

It's the same reason that you can't argue that $\lim_{n \to \infty}(1 + \frac{1}{n})^n = 1$, even though $1 + \frac{1}{n} \to 1$ and $1^n = 1$. The issue is that when we say that the limit of an expression is a certain value, we just mean the expression gets very close to that value - we make no promises about how fast.

When $n$ is very large, $1 - \frac{1}{n}$ is very nearly $1$ - but a number very nearly $1$, when raised to a very large power, can be very small. For example, $0.99^{10000}$ is zero to more than forty decimal places. So that exponent of $n$ can "pull" the value away from one - for $(1 - \frac{1}{n})^n$, it pulls it down, while it pulls $(1 + \frac{1}{n})^n$ up.

Answer 3

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.

Answer 4

Just to give you some intuition, note that since $1-x < 1/(1+x)$ for $0<x<1,$

$$ (1-1/n)^n < \frac{1}{(1 + 1/n)^n}$$

for $n>1.$ Now the right side $\to 1/e.$ Thus if the limit of the left side exists, it has to be $\le 1/e.$