l'Hospital's rule vs limit definition of the derivative

Question

If I have $\lim_{h \to 0} \frac{\ln\left(e+h\right)-1}{h}$, should I recognize this at the limit definition of the derivative where it's $f'\left(e\right)$ where $f\left(x\right)=\ln\left(x\right)$ and get $\frac{1}{e}$, or should I use l'Hopital's rule's repeatedly until I get $\frac{1}{e}$

The reason I ask is because I don't want to be engage in circular logic as how many questions involving $\lim_{x \to 0} \frac{\sin\left(x\right)}{x}$ seem to be concerned about.

Answer 1

Well, you could use l'Hôpital's rule – one application gets you $\lim_{h\to0}\frac{1/(e+h)}1$ – but recognising the derivative form is faster.

Answer 2

In this case the best way is to use the definition of derivative which is more elementary with respect to l'Hospital's rule which can be useful for limits not directly solveable by other elementary methods as for example

$$\lim_{h \to 0} \frac{\ln\left(1+h\right)-h}{h^2}$$

Answer 3

If you need to be more rigorous than simply recognizing that the pattern matches the limit definition of the derivative, my approach would be to define a function $$f(x)=\ln x$$ then upon proving via limits \begin{align} f'(x)&=\lim_{h\to0}\frac{\ln(x+h)-\ln x}{h}\\ &=\lim_{h\to0}\frac1h\ln\left(1+\frac{h}{x}\right)\\ &\;\;\vdots\\ &=\frac1x \end{align} we can assert that the limit you are asked to evaluate is equivalent to $f'(e)=\dfrac1e$.