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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Michael Rozenberg
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Dave
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if d'hospital is not allowed use taylor expansion. – Max Jan 08 '17 at 12:51
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I can't use L'Hôpital's rule and Taylor series. – Dave Jan 08 '17 at 12:51
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@divisor When $a \to 0$, $\textrm{ln}(1+a) \sim a$. – bing Jan 08 '17 at 12:52
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Yes, but it is: $\frac{0}{0}$ ? – Dave Jan 08 '17 at 12:53
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plug in the taylor expansion and cancel out $a$. – Max Jan 08 '17 at 12:55
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@Max without Taylor expansion. – Dave Jan 08 '17 at 12:58
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What definition do you use for the logarithm function? – Jan 08 '17 at 13:18
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http://images.slideplayer.com/10/2810060/slides/slide_20.jpg – Dave Jan 08 '17 at 13:26
4 Answers
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Because $\ln$ is a continuous function.
Thus, $\lim\limits_{a\rightarrow0}\frac{\ln(1+a)}{a}=\ln\lim\limits_{a\rightarrow0}(1+a)^{\frac{1}{a}}=\ln{e}=1$
Michael Rozenberg
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$\ln\lim\limits_{a\rightarrow0^+}(1+a)^{\frac{1}{a}}$; $(1+a)^{\frac{1}{a}}=1^{\frac{1}{0}}$? – Dave Jan 08 '17 at 13:01
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@divisor $\lim\limits_{a\rightarrow0^+}(1+a)^{\frac{1}{a}}=e$. It follows from $\lim\limits_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n=e$. – Michael Rozenberg Jan 08 '17 at 13:03
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@divisor Take a calculator and calculate $\left(1+\frac{1}{1000000}\right)^{1000000}$. – Michael Rozenberg Jan 08 '17 at 13:07
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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:
$$f'(0)=\left.\frac1{1+a}\right|_{a=0}=1$$
DonAntonio
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Hint: $$\ln(1+x)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}nx^n$$
idk
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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 – idk Jan 08 '17 at 13:14
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$\frac{\ln(1+x)}{x}=\sum_{n=1}^\infty\frac{(-1)^{n+1}}nx^{n-1}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\frac{x^4}{5}+...$ – idk Jan 08 '17 at 13:26
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Yes,it is. See more here http://mathworld.wolfram.com/MaclaurinSeries.html – idk Jan 08 '17 at 13:30
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Your solution is correct, but I can't use it. I upvote this, thank you and sorry :( – Dave Jan 08 '17 at 13:34
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You can't just evaluate the function at $0$, because it's not even defined there. Anyway you can get:
$$\lim_{a \to 0} \frac{\ln(1+a)}{a} = \lim_{a \to 0} \frac{\ln(1+a) - \ln(1)}{(a+1) - 1} = (\ln(1+x))'|_{x=0} = 1$$
Stefan4024
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