Determination of $1^\infty$ indeterminate forms Recently I have been learning some of the basic concepts of limits and in my academics. There I have been taught some methods to evaluate indeterminant forms like $1^\infty$, $0^0$ and $\infty^0$.
Like one of the method to evaluate $1^\infty$ form is right here,

$$\lim_{x\to a}f(x)^{g(x)}=\exp(\lim_{x\to a} g(x)[f(x)-1])$$

where $\lim_{x\to a}f(x)=1$ and $\lim_{x\to a}g(x)=\infty$
I also tried some of weary examples that seem to be impossible without this method. Here is one example.

Example 1: Evaluate
$$\lim_{x\to \infty}\left( \frac{p^\frac{1}{x}+q^\frac{1}{x}+r^\frac{1}{x}+s^\frac{1}{x}}{4}\right)^{3x}$$
Answer: First I replaced $\frac{1}{x}\rightarrow z$ equation becomes
$$\lim_{z\to 0}\left( \frac{p^{z}+q^{z}+r^{z}+s^{z}}{4}\right)^\frac{3}{z}$$
which is a $1^\infty$ form
furthur applying the method,
$$= \exp\left(\lim_{z\to 0}\left( \frac{p^{z}+q^{z}+r^{z}+s^{z}}{4}-1\right)\cdot \frac{3}{z}\right)$$
$$= \exp\left(\lim_{z\to 0}\left( \frac{(p^{z}-1)+(q^{z}-1)+(r^{z}-1)+(s^{z}-1)}{z}\right)\cdot \frac{3}{4}\right)$$
Now we know $\lim_{x\to 0}\frac{a^x-1}{x}=\ln a$. Hence,
$$= \exp\left((\ln p+\ln q+\ln r+\ln s)\cdot \frac{3}{4}\right)$$
$$= \exp\left(\ln pqrs\cdot \frac{3}{4}\right)$$
$$= pqrs^{\frac{3}{4}}$$

Now my question, I had perfectly understood that how we have to apply the formula to simplify the indeterminant but I still didn't know that how this formula came to existence. I really didn't know anything about it's proof and that's the thing which I have been searching here.
Before answering the question please be informed that I am a beginner in calculus so please make it as easy as you can. If you think that it's quite difficult to explain this in easy way then you can of course go with the usual one, I'll try my best to understand it and if necessary I'll show it to my teacher to understand it.
Thanks in advance!

Proof:
We'll use a known inequality:
$$
1-\frac{1}{x} < \ln x < x-1 \ \ \ \text{ for } \ \ \ x > 0
$$
Now
$$
\lim_{x\to a}[f(x)]^{g(x)} = \lim_{x\to a} \exp (g(x) \ln f(x) )
$$
Since exp is continuous it will suffice if we will examine the limit of its argument:
$$
\lim_{x\to a} g(x) \ln f(x) \leq \lim_{x\to a} g(x) (f(x) - 1)
$$
And on the other side:
$$
\lim_{x\to a} g(x) \ln f(x) \geq \lim_{x\to a} g(x)\left[ \frac{f(x) - 1}{f(x)}\right]
$$
But since $\lim_{x\to a} f(x) = 1$, we have:
$$
\lim_{x\to a} g(x)\left[ \frac{f(x) - 1}{f(x)}\right] = \lim_{x\to a} g(x) (f(x) - 1)
$$
Thus:
$$
\exp\left(\lim_{x\to a} g(x) (f(x) - 1)\right) \leq \lim_{x\to a} \exp \left( g(x) \ln f(x) \right) \leq \exp\left(\lim_{x\to a} g(x) (f(x) - 1)\right)
$$
Thus:
$$
\lim_{x\to a} \exp \left( g(x) \ln f(x) \right) = \exp\left(\lim_{x\to a} g(x) [f(x) - 1]\right) \square
$$
