(Supposed) Algebraic proof that $\frac{d}{dx}(e^x)=e^x$ using the limit definition of a derivative

Question

I would like to preface this post by saying that I have not taken calculus formally in school yet, so this is purely what I have read and understood from books and self-study. Also, I believe I made a mistake somewhere in the steps I took while doing this that I would appreciate being pointed out. In advance, thank you!

While re-reading a textbook, I re-learned about the constant "$e$" (which I had previously given no thought to) and, consequently, the function $e^x$. I was bored at the time, so I decided to try and derive the reasoning for $\\\frac{d}{dx}(e^x)=e^x$ from the limit definition of a derivative: $\frac{d}{dx} f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.

I started by inputting $e^x$ into the formula: $$\lim_{h \to 0} \frac{e^{x+h} - e^x}{h}$$ and multiplying it by $\\\frac{h}{e^h-1}$ to cancel out the h (this is where I may have made a mistake): $$\lim_{h \to 0} \frac{e^{x+h} - e^x}{h}\cdot\frac{h}{e^h-1}=\lim_{h \to 0} \frac{e^{x+h} - e^x(h)}{h({e^h-1})}$$ after simplifying (canceling the h's) this you get: $$\lim_{h \to 0} \frac{e^{x+h} - e^x}{{e^h-1}}$$ then I expanded the ${e^x}^{+h}$ and factored it (also possible mistake): $$\lim_{h \to 0} \frac{e^x\cdot e^{h} - e^x}{{e^h-1}}=\lim_{h \to 0} \frac{e^x( e^{h}-1)}{{e^h-1}}$$ the $(e^h-1)$ cancels with the $e^h-1$, which obviously ends as: $$\lim_{h \to 0}e^x =e^x$$ Now, this makes sense to me as it is, but there is definitely a huge chance that this is wrong and that there is an exponential, fractional, algebraic, or arithmetic error somewhere.

If you find a simple mistake, please point it out without being mean, and don't lecture me for exceeding expectations at my age. Again, to whomever is reading this, much appreciated. Thank you!

To clarify this was just a shot at seeing if I could find a way to use the definition to prove that d/dx(e^x)=e^x without using series or sums or anything of that sort.

Answer 1

Traditionally, we don't divide by factors, that go to zero together with h. If you know already the translation formula

$$e^{x+h} - e ^x= e^ x( e^h-1)$$

then you only need to show that $\lim_{h->0}\frac{e^h-1}{h}=1$

that is, to know that $e^0=1, {e^x}'(0)=1$

But that is of course not a proof.

A purely algebraic proof goes by

$$\log' x = \frac{1}{x}$$ $$ \log(x)=\int_1^x \frac{dx}{x} \quad x>0$$ $$\exp(x) = \log^{-1}(x) $$ $$\frac{d}{dx} \ \log(\exp x) \ = \ \frac{d}{dx} x = \ 1 \ = \log'(\exp(x))\ \exp'(x) \ = \ \frac{1}{\exp(x)}\ \exp'(x )$$

Answer 2

I think this one could be you are looking for: You Have multiplied the original function by h/(e^h-1) but we can never be sure that it 'will' be equal to 1. As you are reading calculus I assume you could have come up to 'Indeterminant forms'. Since it is an 'indeterminant form'. Although As numerator and denominator both respectively tends to 0, we can never be sure that both the values will be equal. Speaking in layman language:

" Take an example one value is 0.0000001 and other value is 0.0000000001. They both are nearly zero as we can assume(THIS IS ONLY AN EXAMPLE), but the ratio of other with respect to first is 1000." Therefore you could have multiplied there by a thousand or a million instead of 1, who knows. But as you know if we a multiply an expression by a NON-UNIT value. it's value will definitely differ.

In short, we can not be sure of indeterminant forms as they are not clearly defined at a given point.

But how did you get the correct answer? well, I think it is a mere coincidence. you took those values by which terms in the given expression cancel out and you end up with the right answer. Look at that when you wrote, **"multiplying it by value to cancel out the h."**You intentionally took those values which ended you up having a probable answer.

I hope you get the point. If you have any disagreement, feel free to discuss.