How does $$ \frac{1}{\log(1+\frac{d}{a})} $$ simplify to $\frac{a}{d}$ in the limit that $d$ is tiny relative to $a$? I was wondering if a Taylor expansion would work though I am not sure how to work it.
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Here's a non-Taylor series approach. Consider the denominator $\log(1+\frac{d}{a})=\log(a+d)-\log(a)$, which goes to zero as $d\to0$. This should remind you of the definition of the derivative: indeed,
$$\lim_{d\to 0}\dfrac{\log(a+d)-\log(a)}{d}=\dfrac{d}{dx}\log(x)|_{x=a}=\frac{1}{a}$$
Rearranging, we conclude that for small $d$ we indeed have $1/\log(1+\frac{d}{a})\approx a/d$.
Semiclassical
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How did you rearrange? – user35687 Oct 02 '14 at 18:43
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@user35687: I recombined the numerator to $\log(1+d/a)$, and then took the reciprocal overall. So $\lim\limits_{d\to 0}\dfrac{d}{\log(1+d/a)}=a$, which implies $1/\log(1+d/a)\approx a/d$ for small $d$. – Semiclassical Oct 02 '14 at 18:45
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Are you allowed to take the reciprocal inside the limit? i.e. does 1/(lim 1/x) = lim x as x approaches zero? – user35687 Oct 02 '14 at 18:56
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Yes. It's just the Division law of limits with the numerator function as just $f(x)=1$. (Note that the limit of the denominator is finite and nonzero.) – Semiclassical Oct 02 '14 at 19:42
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$\ln(1+\epsilon)=\epsilon+O(\epsilon^2)$. Or, another way of saying it:
$\ln(1+\epsilon)\approx\epsilon$ when $\epsilon$ is small. Since $\frac da$ is small, we have:
$$\frac1{\log(1+\frac da)}\approx\frac1{\frac da}=\frac ad$$
Akiva Weinberger
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When $\epsilon$ is small, $\epsilon^2$ is tiny, even compared to $\epsilon$! That is, we can ignore anything like $\epsilon^2$, $\epsilon^3$, etc., for the most part. So, I'm basically claiming that the difference between $\ln(1+\epsilon)$ and $\epsilon$ is small compared to $\epsilon$. – Akiva Weinberger Oct 02 '14 at 02:02
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Expand f(x) = ln(1 + x) as a Taylor series around x = 0.
f(0) = 0, f'(0) = 1/(1 + 0) = 1
ln(1 + x) = 0 + x + ... = x.
Substitute d/a for x.
user35687
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