Confusion over $a_k = (1+\frac{1}{k})^k = e$

Question

I'm currently just watching an MIT lecture about differentiating exps and logs and it mentions that $a_k = (1+\frac{1}{k})^k = e.$

I've seen the proof and "understand" that $\lim_{k \to \infty} \ln(a_k) = 1$ s0 the limit of $a_k = e$.

The problem is when I read $e = \lim_{k \to \infty} (1+\frac{1}{k})^k$, it looks like it says $e = 1$ because I read it as $e = (1+(0))^\infty = 1$, and I'm having a hard time working out where the confusion is.

Thanks.

Stating $a_k=e$ is quite different from stating $\lim_{k\to +\infty}a_k = e$. — Jack D'Aurizio, Dec 30 '16 at 21:17
I think you may want to revisit the definition of a limit. Simple plug and play can with "$\infty$" will more than likely be misleading — jameselmore, Dec 30 '16 at 21:20
This might be helpful: http://math.stackexchange.com/q/2067849/9464 — , Dec 30 '16 at 21:26
Thanks guys. The indeterminate explanation cleared things up for me, appreciated. — Ryan Tandy, Dec 30 '16 at 21:31

score 4 · Accepted Answer · answered Dec 30 '16 at 21:35

Note that $\lim_{k\to\infty}\left(1+\frac1k\right)^k$ is a limit of the indeterminate form $1^\infty$. If $k=100$, then $1+\frac1k =1.01$ and $1.01^{100}\approx 2.70481382942153$.

As mentioned in the OP, we can exploit the continuity of the exponential function and write

$$\lim_{k\to\infty}\left(1+\frac1k\right)^k=\lim_{k\to\infty}e^{\log\left(1+\frac1k\right)^k}=e^{\lim_{k\to\infty}\log\left(1+\frac1k\right)^k}=e^{\lim_{k\to\infty}k\log\left(1+\frac1k\right)}\tag 1$$

Then, recalling that $\frac{1}{k+1}\le\log\left(1+\frac1k\right)\le \frac1k$, we arrive at the anticipated result.

Note that in $(1)$, we transformed the indeterminate form $1^\infty$ into the indeterminate form $e^{\infty \times 0}$.

Confusion over $a_k = (1+\frac{1}{k})^k = e$

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