Why $\lim\limits_{a\to 0} \frac{\ln(1+a)}{a}=1$?

Question

In question, to me wrote, that $\lim\limits_{a\to 0} \frac{\ln(1+a)}{a}=1$, but why?

$\lim\limits_{a\to 0} \frac{\ln(1+a)}{a}=\lim\limits_{a\to 0} \frac{\ln(1+0)}{0}=|\frac{0}{0}|$ or am I wrong?

Answer 1

Because $\ln$ is a continuous function.

Thus, $\lim\limits_{a\rightarrow0}\frac{\ln(1+a)}{a}=\ln\lim\limits_{a\rightarrow0}(1+a)^{\frac{1}{a}}=\ln{e}=1$

Answer 2

You can't just evaluate the function at $0$, because it's not even defined there. Anyway you can get:

$$\lim_{a \to 0} \frac{\ln(1+a)}{a} = \lim_{a \to 0} \frac{\ln(1+a) - \ln(1)}{(a+1) - 1} = (\ln(1+x))'|_{x=0} = 1$$

Answer 3

Hint: $$\ln(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}nx^n$$

Answer 4

I suppose you must know by now derivatives, so:

$$f(a):=\log(1+a)\implies f'(0):=\lim_{a\to0}\frac{f(0+a)-f(0)}a=\lim_{a\to0}\frac{\log(1+a)}a$$

and now just substitute:

$$f'(0)=\left.\frac1{1+a}\right|_{a=0}=1$$