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Chapter 9: Problem 42

Find the limit of the following sequences or determine that the limit does not exist. Verify your result with a graphing utility. $$a_{n}=\frac{\ln n}{n^{1.1}}$$

Short Answer

Expert verified
Based on the step-by-step solution provided above, the limit of the sequence as n goes to infinity, defined as $$\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{\ln n}{n^{1.1}}$$, is found to be 0. This result is verified by analyzing the behavior of the sequence, applying L'H么pital's Rule, and evaluating the limit of the ratio of the derivatives. The graphical representation of the sequence confirms this result, as the points of the sequence approach the horizontal line at y=0 as n increases.

Step by step solution

01

Analyze the sequence

As n goes to infinity, we have two competing functions in the sequence: the natural logarithm of n, which grows at a slower rate than any power of n, and n raised to the power of 1.1, which grows much faster than the logarithm. Our task is to determine the limit of the sequence: $$\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{\ln n}{n^{1.1}}$$
02

Apply L'H么pital's Rule

L'H么pital's Rule is used to evaluate the limit of a ratio of functions when the limit of the denominator and numerator goes to 0 or infinity, i.e., it is an indeterminate form. Since both the numerator and denominator go to infinity as n goes to infinity, we can apply L麓H么pital's Rule. To use L'H么pital's Rule, we first find the derivatives of the numerator and the denominator with respect to n: $$\frac{d}{dn}(\ln n) = \frac{1}{n}$$ $$\frac{d}{dn}(n^{1.1}) = 1.1n^{0.1}$$ Now, apply L'H么pital's Rule by finding the limit of the ratio of these derivatives: $$\lim_{n\to\infty} \frac{\frac{d}{dn}(\ln n)}{\frac{d}{dn}(n^{1.1})} = \lim_{n\to\infty} \frac{\frac{1}{n}}{1.1n^{0.1}}$$
03

Evaluate the limit

We can simplify the expression inside the limit: $$\lim_{n\to\infty} \frac{1}{1.1n^{1.1}}$$ As n goes to infinity, the denominator, \(1.1n^{1.1}\), goes to infinity as well, making the entire fraction go to 0. Therefore, the limit of the sequence as n goes to infinity is 0. So, the limit of the sequence is: $$\lim_{n\to\infty} a_n = 0$$
04

Verify the result graphically

Using a graphing utility (such as Desmos or Geogebra), plot the sequence along with a horizontal line at y=0. As n increases, the points of the sequence should approach the horizontal line at y=0, confirming the limit found in Step 3.

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Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty} x^{2 k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

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