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How would I prove that ln (lim (u)) = lim (ln (u))by MadViolinist
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#1
Feb1213, 05:01 PM

P: 18

We were working with L'hospital's rule and my teacher said that the teacher before him told him that this was true:
lim (x→∞) [ln u] = ln ( lim (x→∞) u), where u is a continuous function. My teacher has never found a proof for this, although it works every time. Does anyone know how to prove this? Thanks! 


#2
Feb1213, 06:46 PM

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I guess you are implicitly assuming that ##\lim_{x \rightarrow \infty}u(x)## exists, because otherwise the statement doesn't make any sense. So if we put ##L = \lim_{x \rightarrow \infty}u(x)## and ##x_n## is any sequence such that ##x_n \rightarrow \infty## as ##n \rightarrow \infty##, we have a corresponding sequence ##u_n = u(x_n)## such that ##u_n \rightarrow L## as ##n \rightarrow \infty##. We may now write
$$\ln(L) = \ln(\lim_{x \rightarrow \infty}u(x)) = \ln(\lim_{n \rightarrow \infty} u_n) = \lim_{n \rightarrow \infty} \ln(u_n) = \lim_{n \rightarrow \infty} \ln(u(x_n))$$ The third equality holds because ##\ln## is continuous at ##L##. Thus we have established that $$\lim_{n \rightarrow \infty} \ln(u(x_n)) = \ln(L)$$ This is true for any sequence ##x_n \rightarrow \infty##, so we may conclude that $$\lim_{x \rightarrow \infty} \ln(u(x)) = \ln(L)$$. 


#3
Feb1213, 07:00 PM

P: 18

Thanks for the answer!
However, I do not get the step where ln(limn→∞ un)=limn→∞ ( ln(un)). How did you change the order of the natural log and the limit? 


#4
Feb1213, 07:21 PM

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How would I prove that ln (lim (u)) = lim (ln (u))
Assuming that theorem, all you need is the fact that ##\ln## is continuous at ##L##, where ##L## is any real positive number. (By the way, positivity of ##L## is another assumption that needs to be added to the problem statement, otherwise the equation makes no sense.) How to prove that ##\ln## is continuous depends on how you defined ##\ln##. One standard definition is $$\ln(x) = \int_{1}^{x} \frac{1}{t} dt$$ If we use that definition, then $$\begin{align} \ln(x+h)  \ln(x) &= \left\int_{1}^{x+h} \frac{1}{t} dt  \int_{1}^{x} \frac{1}{t}\right\\ &= \left\int_{x}^{x+h}\frac{1}{t} dt\right\\ \end{align}$$ If ##h > 0## then we have the bound $$\left\int_{x}^{x+h}\frac{1}{t} dt\right \leq \left\int_{x}^{x+h}\frac{1}{x} dt\right = \left \frac{h}{x} \right$$ which we can make as small as we like as ##h \rightarrow 0##. A similar argument holds for ##h < 0##. 


#5
Feb1213, 07:27 PM

P: 18

Thanks again!



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