How to find the limit of a logarithm when doing infinite series'
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Dr.Peterson
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What does the expression inside the limit equal? (Expand it as a difference of two expressions.)
The first expression approaches ln(1) = 0; the second is a constant, ln(1/2) = -ln(2).
What is the result of the subtraction?
Show us your thinking, so we can identify any errors in your work.
The first expression approaches ln(1) = 0; the second is a constant, ln(1/2) = -ln(2).
What is the result of the subtraction?
Show us your thinking, so we can identify any errors in your work.
Expanding this gets: [MATH]\lim_{a\to\infty}\ln(\frac{a}{a+1})-\ln(\frac12)[/MATH]. [MATH]-\ln(\frac12)=\ln(2)[/MATH]. I wasn't sure how the rest worked, but now, as I'm writing this out again I see that the [MATH]\lim_{a\to\infty}\ln(\frac{a}{a+1})=ln(1)[/MATH] - which can be seen either with l'hopitals rule or just by realizing that the x's in the fraction will dominate. Now I understand! Thank you!
Dr.Peterson
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Exactly. And this is part of the reason we ask to see your work - so you can realize you know how!
Another way to deal with [MATH]\frac{a}{a+1}[/MATH] is to carry out the division: [MATH]\frac{a}{a+1} = 1 - \frac{1}{a+1}[/MATH].
Another way to deal with [MATH]\frac{a}{a+1}[/MATH] is to carry out the division: [MATH]\frac{a}{a+1} = 1 - \frac{1}{a+1}[/MATH].
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If I were faced with the limit:
[MATH]L=\lim_{a\to\infty}\left(\left.\ln\left(\frac{x}{x+1}\right)\right|_1^a\right)[/MATH]
I would work within the parentheses on the limit first:
[MATH]L=\lim_{a\to\infty}\left(\ln\left(\frac{2a}{a+1}\right)\right)[/MATH]
[MATH]L=\ln\left(\lim_{a\to\infty}\left(\frac{2}{1+\dfrac{1}{a}}\right)\right)=\ln(2)[/MATH]
[MATH]L=\lim_{a\to\infty}\left(\left.\ln\left(\frac{x}{x+1}\right)\right|_1^a\right)[/MATH]
I would work within the parentheses on the limit first:
[MATH]L=\lim_{a\to\infty}\left(\ln\left(\frac{2a}{a+1}\right)\right)[/MATH]
[MATH]L=\ln\left(\lim_{a\to\infty}\left(\frac{2}{1+\dfrac{1}{a}}\right)\right)=\ln(2)[/MATH]
HallsofIvy
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Your question is "How does \(\displaystyle \lim_{a\to \infty}\left[ln\left(\frac{x}{x+ 1}\right)\right]_1^a= 0.69\). The answer to that is "it doesn't!". That limit is ln(2) which is approximately 0.69.
In the first place, \(\displaystyle \left[ln\left(\frac{x}{x+ 1}\right)\right]_1^a\) means \(\displaystyle ln\left(\frac{a}{a+1}\right)- \ln\left(\frac{1}{1+1}\right)= ln\left(\frac{2a}{a+1}\right)\). Further, ln(x) is continuous for all positive x so this limit is the same as \(\displaystyle ln\left(\lim_{a \to\infty}\frac{2a}{a+ 1}\right)\).
To find \(\displaystyle \lim_{a \to\infty}\frac{2a}{a+ 1}\), divide both numerator and denominator by a: \(\displaystyle \frac{2a}{a+ 1}= \frac{2}{1+\frac{1}{a}}\) and, now, as a goes to infinity, 1/a goes to 0 so that limit is \(\displaystyle \frac{2}{1}= 2\). \(\displaystyle ln\left(\lim_{a \to\infty}\frac{2a}{a+ 1}\right)= ln(2)\).
In the first place, \(\displaystyle \left[ln\left(\frac{x}{x+ 1}\right)\right]_1^a\) means \(\displaystyle ln\left(\frac{a}{a+1}\right)- \ln\left(\frac{1}{1+1}\right)= ln\left(\frac{2a}{a+1}\right)\). Further, ln(x) is continuous for all positive x so this limit is the same as \(\displaystyle ln\left(\lim_{a \to\infty}\frac{2a}{a+ 1}\right)\).
To find \(\displaystyle \lim_{a \to\infty}\frac{2a}{a+ 1}\), divide both numerator and denominator by a: \(\displaystyle \frac{2a}{a+ 1}= \frac{2}{1+\frac{1}{a}}\) and, now, as a goes to infinity, 1/a goes to 0 so that limit is \(\displaystyle \frac{2}{1}= 2\). \(\displaystyle ln\left(\lim_{a \to\infty}\frac{2a}{a+ 1}\right)= ln(2)\).