Is the formula $\lim\limits_{x\to a} (1+f(x))^{g(x)}=e^{\lim\limits_{x\to a}f(x)g(x)}$ a standard result?

Is the formula for the form $1^{\infty}$ that is $$\lim\limits_{x\to a} (1+f(x))^{g(x)}=e^{\lim\limits_{x\to a}f(x)g(x)}$$ a standard formula? Does it have any special name? Note:In the above formula $f(x)\to 0$ as $x\to a$ and $g(x)\to\infty$ as $x\to a$

I'm asking because I'm not sure whether it will be allowed in a subjective maths exam(it sometimes greatly simplifies calculations).Thanks.

One of the logarithmic inequalities says: $$\frac{x}{1+x} < \ln (1 + x) < x, \forall x > -1$$

Because $\lim\limits_{x\rightarrow a} f(x)=0$, then $f(x)>-1$, for all $x$ in some vicinity of $a$. So $$\frac{f(x)}{1+f(x)} < \ln (1 + f(x)) < f(x)$$ from all $x$ in some vicinity of $a$. As a result $$\lim_{x \rightarrow a} \left ( 1 + f(x) \right )^{\frac{1}{f(x)}}=e$$

Alternatively (let's limit to the case when $g(x)>0$) $$\frac{f(x) \cdot g(x)}{1+f(x)} < g(x) \cdot \ln (1 + f(x)) < f(x)\cdot g(x)$$ ($e^{x}$ is ascending $\forall x$) $$e^{\frac{f(x) \cdot g(x)}{1+f(x)}} < (1 + f(x))^{g(x)} < e^{f(x)\cdot g(x)}$$ Or $$\frac{e^{\frac{f(x) \cdot g(x)}{1+f(x)}}}{e^{f(x)\cdot g(x)}} < \frac{(1 + f(x))^{g(x)}}{e^{f(x)\cdot g(x)}}<1$$ Or $$0 < \frac{1}{e^{f(x) \cdot g(x) \cdot \frac{f(x)}{1+f(x)}}} < \frac{(1 + f(x))^{g(x)}}{e^{f(x)\cdot g(x)}}<1 \tag{*}$$ Now a little summary:

• if $g(x)$ is bounded (at least in some vicinity of $a$), then $\mathbf{(*)}$ is asymptotic to $1$ when $x\rightarrow a$.
• same is true if $g(x)$ happens to be continuous in a vicinity of $a$.

Not totally sure about the $\lim\limits_{x\rightarrow a} g(x)=\infty $ case. I would say, it depends.