How do I prove
$\lim\limits_{a \to \infty} \ln((\frac{a+n}{a})^a) = n$
How do I prove $\lim\limits_{a \to \infty} \ln((\frac{a+n}{a})^a) = n$
-1
$\begingroup$
limits
logarithms
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0You have a conflict with your variables (presumably the limit variable is $a$). – 2017-01-03
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0Yes sorry, corrected now. – 2017-01-03
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0Are you familiar with limits of the form: $\lim_{x\rightarrow x_0}\left(1+f(x)^{-1}\right)^{f(x)}$, where $f(x)\rightarrow\infty$ as $x\rightarrow x_0$? – 2017-01-03
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0I wonder what kind of text would choose, in a question like this one, precisely $\;a\;$ to be the variable and $\;n\;$ a constant... – 2017-01-03
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0@DonAntonio I made it up myself. – 2017-01-03
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0@SomeStrangeUser No I am not. – 2017-01-03
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0See https://en.wikipedia.org/wiki/E_(mathematical_constant)#Representations – 2017-01-03
3 Answers
2
We have :$$\lim_{a \to \infty} \ln\left(\frac{a+n}{a}\right)^a = \lim_{a \to \infty} a \ln\left(1+ \frac{n}{a}\right) = \lim_{a \to \infty} \frac{\ln\left(1+ \frac{n}{a}\right)}{\frac{n}{a}} \times n = n$$
0
$$\ln\left(\left(\frac{a+n}{a}\right)^a\right)=a\ln\left(1+\frac{n}{a}\right)$$
Use the fact that $\ln(1+x)=x+o(x)$ when $x$ is small and you'll get the result.
0
$\lim_\limits{a\to\infty}(1+\frac 1a)^a = e\\ \lim_\limits{a\to\infty}(1+\frac 1a)^na = e^n\\ \lim_\limits{a\to\infty}((1+\frac 1a)^n)a = e^n\\ \lim_\limits{a\to\infty}(1+\frac na)^a = e^n\\ \lim_\limits{a\to\infty}(\frac {a+n}a)^a = e^n\\ \lim_\limits{a\to\infty}\ln(\frac {a+n}a)^a = n$