The Generalisation of $\lim _{x\to \:0}\:\frac{\left(e^x-1\right)}{x}=1$

Solution 1:

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.