Is there an intuitive see why $\lim_{n \to \infty} \frac{e^{x/n}-1}{x/n} = 1$

Question

I'm almost certain the following is true:

$$\lim_{n \to \infty} \frac{e^{x/n}-1}{x/n} = 1$$

At first blush, it would seem (to me at least) that this would approach infinity. However, upon further investigation however, it doesn't...it's 1. Is there a some intuitive way to think about this?

Answer 1

The limit in your post is zero, since $e^{x/n} \to 1$ as $n \to \infty$. However, you are probably thinking of something like $$\lim_{n \to \infty} \frac{e^{x/n}-1}{x/n} = 1,$$ which can be viewed roughly as "$e^{x/n} - 1 \approx x/n$" as $n \to \infty$.

Indeed, by replacing $y=x/n$ and taking $y \to 0$, the above is the same as $$\lim_{y \to 0} \frac{e^y-1}{y} = 1.$$ This can be verified by writing out the definition of the derivative of $f(y)=e^y$ at zero: $f'(y) = 1$.

Alternatively, this can be verified by looking at the Taylor series $e^y = 1 + y + O(y^2)$ as $y \to 0$, as mentioned in the comments.

Answer 2

My answer is referring to the original question:

I'm almost certain the following is true:

$\displaystyle \lim_{n \rightarrow \infty} \left(e^{\frac x n} - 1 \right) = {\frac x n}$

At first blush, it would seem (to me at least) that this limit would be 0. However, upon further investigation however, it isn't...rather it's x/n. Is there a some intuitive way to think about this?

It is true that

$$\lim_{n \to \infty}\left(1+\frac{x}n\right)^n=e^x \qquad \color{blue}\checkmark$$

It seems that you then made some transformations which are not valid. You take the n-th root on both sides. But since $n \to \infty$ on the LHS you cannot just put it on the the RHS like

$$\lim_{n \to \infty}1+\frac{x}{n}=e^{x/n} \qquad \color{red}X$$

Consequently subtracting 1 on both sides does not make it even better.

Answer 3

$$ \lim_{\Delta x\to0} \frac{e^{\Delta x} - e^0}{\Delta x - 0} = \lim_{\Delta x \to 0} \frac{\Delta e^x}{\Delta x} = \frac{d e^x}{dx} = e^x = e^x = 1 \text{ if } x=0. $$