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Chapter 7: Problem 9

Evaluate \(\lim _{n \rightarrow \infty} n\left(\ln \left(3+\frac{1}{n}\right)-\ln 3\right)\).

Short Answer

Expert verified
The limit of the given expression is \(-\frac{1}{3}\).

Step by step solution

01

Rewrite the given expression using logarithm properties

Recall that the subtraction of logarithms with the same base can be expressed as a single logarithm whose argument is the ratio of the arguments of the original logarithms. Rewrite the expression using the logarithm properties: \[\lim_{n\to\infty} n\left(\ln\left(3+\frac{1}{n}\right)-\ln 3\right) = \lim_{n\to\infty} n\ln\left(\frac{3+\frac{1}{n}}{3}\right)\]
02

Simplify the expression inside the logarithm

Simplify the expression inside the logarithm by combining the fractions: \[\lim_{n\to\infty} n\ln\left(\frac{3+\frac{1}{n}}{3}\right) = \lim_{n\to\infty} n\ln\left(\frac{3n+1}{3n}\right)\]
03

Rewrite the expression as a quotient

Recall that L'Hopital's rule requires that the limit be in the form \(\lim_{x\to a} \frac{f(x)}{g(x)}\). Rewrite the expression as a quotient using the property that the logarithm of a number raised to a power is equal to the power times the logarithm: \[\lim_{n\to\infty} n\ln\left(\frac{3n+1}{3n}\right) = \lim_{n\to\infty} \frac{\ln\left(\frac{3n+1}{3n}\right)}{\frac{1}{n}}\]
04

Apply L'Hopital's rule

Now that the expression is in the form required to apply L'Hopital's rule, differentiate both the numerator and the denominator with respect to \(n\): \[\lim_{n\to\infty} \frac{\ln\left(\frac{3n+1}{3n}\right)}{\frac{1}{n}} = \lim_{n\to\infty} \frac{\frac{d}{dn}\ln\left(\frac{3n+1}{3n}\right)}{\frac{d}{dn}\frac{1}{n}}\] Using the chain rule, we find the derivative of the numerator: \[\frac{d}{dn}\ln\left(\frac{3n+1}{3n}\right) = \frac{1}{\frac{3n+1}{3n}} \cdot \frac{d}{dn}\left(\frac{3n+1}{3n}\right) = \frac{3n}{3n+1} \cdot \frac{-3}{9n^2}\] And the derivative of the denominator: \[\frac{d}{dn} \frac{1}{n} = -\frac{1}{n^2}\]
05

Simplify the expression and take the limit

Now substitute the derivatives back into the limit: \[\lim_{n\to\infty} \frac{\frac{3n}{3n+1} \cdot \frac{-3}{9n^2}}{-\frac{1}{n^2}}\] Simplify the expression: \[\lim_{n\to\infty} \frac{3n}{3n+1} \cdot \frac{-3n^2}{9n^2} = \lim_{n\to\infty} \left(\frac{3n}{3n+1}\right) \left(\frac{-3}{9}\right)\] Further simplify: \[\lim_{n\to\infty} \left(\frac{3n}{3n+1}\right) \left(-\frac{1}{3}\right) = -\frac{1}{3} \lim_{n\to\infty} \left(\frac{3n}{3n+1}\right)\] Now take the limit as \(n\) approaches infinity: \[-\frac{1}{3} \lim_{n\to\infty} \left(\frac{3n}{3n+1}\right) = -\frac{1}{3} \cdot \frac{\infty}{\infty+1} = -\frac{1}{3}\] So, the limit of the given expression is \(-\frac{1}{3}\).

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Most popular questions from this chapter

Consider an arithmetic sequence with first term b and difference \(d\) between consecutive terms. (a) Write the sequence using the three-dot notation, giving the first four terms of the sequence. (b) Give the \(100^{\text {th }}\) term of the sequence. \(b=0, d=\frac{1}{3}\)

Show that $$ \ln n<1+\frac{1}{2}+\cdots+\frac{1}{n-1} $$ for every integer \(n \geq 2\).

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