I am a curious Calculus student with an interesting question.
The derivative of $e^{x}$ = $e^{x}$, however taking the derivative using limits, and not sums or some other method, is difficult.
Perhaps would taking the limit of e, which is $\lim_{n \to \infty} {(1+{1\over n})^n}$ and the $\lim_{h \to 0} {(e^{x+h} - e^{x})\over h}$, which is the derivative formula.
Now, these limits have different terms, n and h respectively, which go to 0 and infinity. In order to get similar terms, an equivalency of the limits could be set up.
$\lim_{h \to 0} {1\over h} = \infty$, and $\lim_{n \to \infty} n = \infty$
$\lim_{h \to 0} {1\over h}$ = $\lim_{n \to \infty} n$ and vice versa for $1\over n$ to $h$.
Rearranging the equation now gives:
$\lim_{n \to \infty} n({(1+{1\over n})^{n(x+{1\over n})} - (1+{1 \over n})^{nx}})$ or $\lim_{n \to \infty} n({(1+{1\over n})^{nx + 1} - (1+{1 \over n})^{nx}})$
Factoring out $(1+{1 \over n})^{nx}$ gives:
$\lim_{n \to \infty} n((1 + {1\over n})^{nx})((1 + {1 \over n})^1 - 1)$ or $\lim_{n \to \infty} n((1 + {1\over n})^{nx})(1 + {1 \over n} - 1)$
Removing the 1's gives $\lim_{n \to \infty} n((1 + {1\over n})^{nx})({1 \over n})$
Distributing out the n gives:
$\lim_{n \to \infty} (1 + {1\over n})^{nx}$ = $e^x$
This method works for $e^{ax}$ as well, where a is equal to some constant coefficient.
For example: $d\over{dx}$ $e^{3x}$ = $3e^{3x}$.
I will apply the previous method to $e^{3x}$.
Setting up the limit:
$\lim_{n \to \infty} {n}({(1+{1\over n})^{3n(x+{1\over n})} - (1+{1 \over n})^{3nx}})$ or $\lim_{n \to \infty} {n}({(1+{1\over n})^{3nx + 3} - (1+{1 \over n})^{3nx}})$
Factoring out $(1+{1 \over n})^{3nx}$ gives:
$\lim_{n \to \infty} {n}((1+{1\over n})^{3nx})((1+{1\over n})^3 - 1)$
Multiplying out the $(1+{1 \over n})^3$ gives:
$\lim_{n \to \infty} {n}((1+{1\over n})^{3nx})(1+ {3 \over n} + {3 \over n^2} + {1 \over n^3}- 1)$
Subtracting the 1's gives:
$\lim_{n \to \infty} {n}((1+{1\over n})^{3nx})({3 \over n} + {3 \over n^2} + {1 \over n^3})$
Distributing the n gives:
$\lim_{n \to \infty} ((1+{1\over n})^{3nx})(3 + {3 \over n} + {1 \over n^2})$
As n approaches $\infty$, $3 \over n$ and $1 \over n^2$ approach $0$.
With that in mind, and removing the diminishing numbers, gives us:
$\lim_{n \to \infty} 3(1+{1\over n})^{3nx}$ = $3e^{3x}$.
It seems that using this method, the only significant term in the expanded binomial is the second term. Using binomial theorem, the second term of an expanded binomial is $\binom{a}{1} \over n$ which multiplied by the $n$ just becomes $\binom{a}{1}$.
The second term of any combination is just the ${a}$ in $\binom{a}{k}$ where $k$ is some number $0 \le k\le a$.
Thus, for all possible values $a$ in $d \over {dx}$ $e^{ax}$, this method will yield the correct derivative, $a e^{ax}$.
My question is, since this "works", is it even following the rules of mathematics? Nowhere else have I seen this method, I just tried it out one day and saw that it worked. Can limits be exchanged like that?
If all these answers are yes, then has someone does this before, who, and when?
I feel like this insight isn't something new, and it'll probably look elementary, but my math teachers haven't seen this method before, so who knows?
Thanks for looking and responding!
--Additional application for $e^{0x}$--
$\lim_{n \to \infty} n({(1+{1\over n})^{n(0+{1\over n})} - (1+{1 \over n})^{0}})$
which equals:
$\lim_{n \to \infty} n(1 + {1 \over n} - 1)$ = $1$, which is equal to $e^{0x}$ = $e^0$.
This argument is shaky from the very beginning, since it's far from clear why $$ \lim_{h\to 0} \frac{1}{h} \left( \left( \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) \right)^{x+h} - \left( \lim_{n \to \infty}\left( 1 + \frac{1}{n} \right) \right)^{x} \right) $$ would equal $$ \lim_{n \to \infty} \frac{1}{1/n} \left( \left( 1 + \frac{1}{n} \right)^{x+\frac{1}{n}} - \left( 1 + \frac{1}{n} \right)^{x} \right) . $$ To begin with, when you take $h = 1/n$ with $n\to\infty$, you're only considering the one-sided limit $h \to 0^+$, but that's not the serious problem. The real issue is replacing the double limits, where $n \to \infty$ first and then $h \to 0$, with just a single limit.
An example, to show why this need not work: $$ \lim_{h \to 0} \left( \lim_{n \to \infty} \frac{1}{hn} \right) = \lim_{h \to 0} 0 = 0 , $$ but $$ \lim_{n \to \infty} \frac{1}{\frac{1}{n} \, n} = \lim_{n \to \infty} 1 = 1 . $$