The Generalisation of $\lim _{x\to \:0}\:\frac{\left(e^x-1\right)}{x}=1$ $$\lim _{\:f\left(x\right)\to \:0}\left(\frac{\left(e^{f\left(x\right)}-1\right)}{x}\right)$$
Is this equal to one, or this bottom one
$$\lim _{\:f\left(x\right)\to \:0}\left(\frac{\left(e^{f\left(x\right)}-1\right)}{f\left(x\right)}\right)$$
I am pretty sure it is the second one, but my teacher wrote the first one
. As an example, I typed in these two to compare the values in symbolab, but it wouldn't show the answer
$$\lim _{x^3\to \:0}\left(\frac{\left(e^{x^3}-1\right)}{x}\right)$$
$$\lim _{x^3\to \:0}\left(\frac{\left(e^{x^3}-1\right)}{x^3}\right)$$
I used the expansion of e and got the first answer as zero and the second as $1$. Is it correct?
 A: You are essentially right, but also the second one requires caution because
$$\lim _{\:f\left(x\right)\to \:0}\left(\frac{\left(e^{f\left(x\right)}-1\right)}{f\left(x\right)}\right)$$
is not really a "proper" limit. What does $f(x) \to 0$ mean? You should not regard $f(x)$ as an indendent variable $y$, it may happen that $\lvert f(x) \rvert \ge r > 0$ in which case it is impossible that $f(x) \to 0$. And if you consider
$$\lim _{\:x \to \:0}\left(\frac{\left(e^{f\left(x\right)}-1\right)}{f\left(x\right)}\right) ,$$
you have the problem that the quotient is not defined for $f(x) = 0$.
The first one is definitely false in general. Even if you would find a reasonable  interpretation of $f(x) \to 0$, it is not clear that $x \to 0$ as $f(x) \to 0$. You may consider
$$\lim _{\:x \to \:0}\left(\frac{\left(e^{f\left(x\right)}-1\right)}{x}\right) , \tag{1}$$
but in general this limit will not exist. The minimal requirement is that $f(x) \to 0$ as $x \to 0$. So let us assume that  $f(x) \to 0$ as $x \to 0$ and that $f$ is differentiable. Then by L'Hospital's rule we may consider
$$\lim _{\:x \to \:0}\left(\frac{f'(x)e^{f(x)}}{1}\right) . \tag{2}$$
It this limit exists, then also the limit $(1)$ exists and equals the limit $(2)$. But you see that the limit $(2)$ exists iff $\lim _{\:x \to \:0}f'(x)$ exists. This is not necessarily true, and even if it exists, it may have any value.
A: It is certainly more practical to work with the $o()$ notation.
$\lim\limits_{x\to 0}\dfrac{e^x-1}{x}=1\iff e^x=1+x+o(x)$  in a neighbourhood of zero.
So as long as $f(x)\to 0$ we can rewrite
$\dfrac{e^{f(x)}-1}{g(x)}=\dfrac{1+f(x)+o(f(x))-1}{g(x)}=\dfrac{f(x)+o(f(x))}{g(x)}\sim \dfrac{f(x)}{g(x)}$
The problem is reduced to study the limit of $f/g$ in zero.
