Why $\lim_{n\to \infty} \left(1+ \frac 1n \right)^n=e$ doesn't imply $\lim_{n\to \infty} \left(1+\frac 1n \right)≠1?$

Question

I'm looking for the answer to this question. But I could not find the "satisfactory" answer.

This is obvious,

$$\lim_{n\to \infty} \left(1+\frac 1n \right)=1+0=1$$

and

$$\lim_{n\to \infty} \left(1+ \frac 1n \right)^n=e≈2.718281...$$

And also we can write,

$$\lim_{n\to \infty} \left(1+ \frac 1n \right)^n=\left(1+0 \right)^{\infty}=1^{\infty}≠1$$

The point, which that I can not understand is:

$$\lim_{n\to \infty} 1^{\infty}=1^{\infty}=1$$

But, this is also true:

$$\lim_{n\to \infty} \left (1+\frac 1n \right) =\lim_{n\to\infty}1=1$$

Why $\lim_{n\to \infty} \left(1+ \frac 1n \right)^n=e$ doesn't imply $\lim_{n\to \infty} \left(1+\frac 1n \right)≠1?$

Answer 1

To see or remember why $1^{\infty}$ is an indeterminate for just consider

• $f(x)\to 1$
• $g(x) \to \infty$

then

$$f(x)^{g(x)}=e^{g(x)\log f(x)}$$

and

$$g(x)\log f(x)$$

is an ideterminate form $\infty \cdot 0$, but if $f(x)=1$ then $1^{g(x)}=1$.

Answer 2

To address the question in the title, the premise $\lim_{n\to \infty} \left(1+ \frac 1n \right)^n=e$ is true and the conclusion $\lim_{n\to \infty} \left(1+\frac 1n \right) \ne 1$ is false, so the implication is false.

As the comments point out, the source of error is that $1^\infty$ is an indeterminate form like $\dfrac{0}{0}$, so you can manipulate it as you like, say.

$$\left( \underbrace{{\huge 1} + {\small 1/n}}_\mbox{tends to 1} \right)^{n \cdot \color{blue}{2018}} \to e^{\color{blue}{2018}}$$

You can change $\color{blue}{2018}$ to any number you like, so $1^\infty$ can't even be defined.