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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?

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    if d'hospital is not allowed use taylor expansion. – 2017-01-08
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    I can't use L'Hôpital's rule and Taylor series. – 2017-01-08
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    @divisor When $a \to 0$, $\textrm{ln}(1+a) \sim a$. – 2017-01-08
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    Yes, but it is: $\frac{0}{0}$ ? – 2017-01-08
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    plug in the taylor expansion and cancel out $a$. – 2017-01-08
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    @Max without Taylor expansion. – 2017-01-08
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    What definition do you use for the logarithm function? – 2017-01-08
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    http://images.slideplayer.com/10/2810060/slides/slide_20.jpg – 2017-01-08

4 Answers 4

6

Because $\ln$ is a continuous function.

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

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    $\ln\lim\limits_{a\rightarrow0^+}(1+a)^{\frac{1}{a}}$; $(1+a)^{\frac{1}{a}}=1^{\frac{1}{0}}$? – 2017-01-08
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    @divisor $\lim\limits_{a\rightarrow0^+}(1+a)^{\frac{1}{a}}=e$. It follows from $\lim\limits_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n=e$. – 2017-01-08
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    Michael Rozenberg, why is that? – 2017-01-08
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    And why $a\rightarrow0^+$, not $a\rightarrow0$? – 2017-01-08
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    @divisor Take a calculator and calculate $\left(1+\frac{1}{1000000}\right)^{1000000}$. – 2017-01-08
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    @divisor $a\rightarrow0^+$ was typo. Thank you! – 2017-01-08
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    I understand $(1+\frac{1}{1000000}$, but $z^{1000000}$? – 2017-01-08
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51399/discussion-between-divisor-and-michael-rozenberg). – 2017-01-08
5

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:

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

3

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

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    $\lim_{a\to 0} \ln(1+a) = 0 $ ? – 2017-01-08
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    $\ln(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}nx^n$ so $\frac{\ln(1+x)}{x}=\sum_{n=1}^\infty\frac{(-1)^{n+1}}nx^{n-1}$ now expands the series and take the limit $x\rightarrow 0$. Note that the first series term will be 1 and the others 0 – 2017-01-08
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    Is there a simpler explanation? – 2017-01-08
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    $\frac{\ln(1+x)}{x}=\sum_{n=1}^\infty\frac{(-1)^{n+1}}nx^{n-1}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\frac{x^4}{5}+...$ – 2017-01-08
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    It is Taylor series? – 2017-01-08
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    Yes,it is. See more here http://mathworld.wolfram.com/MaclaurinSeries.html – 2017-01-08
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    Your solution is correct, but I can't use it. I upvote this, thank you and sorry :( – 2017-01-08
1

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

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

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    This a derivative? – 2017-01-08
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    @divisor Yeah it's the definition for derivative – 2017-01-08
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    Unavailable for me... – 2017-01-08