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Chapter 8: Problem 71

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$

Short Answer

Expert verified
Question: Using the formal definition of the limit, prove that $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$. Answer: To prove the limit using the formal definition, we need to find a positive integer \(N\) such that for any \(\epsilon > 0\), it holds that for all \(n > N\), $$\left|\frac{3 n^{2}}{4 n^{2}+1} - \frac{3}{4}\right| < \epsilon$$. After simplifying the expression within the absolute value, we found that $$N = \left\lceil\sqrt{\frac{3}{16\epsilon}}\right\rceil$$ satisfies the given condition. Therefore, by the formal definition of the limit, we have proved that $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$.

Step by step solution

01

Rewrite the expression

Start by rewriting the expression as follows: $$\frac{3 n^{2}}{4 n^{2}+1} - \frac{3}{4} = \frac{3n^2 - (3/4)(4n^2+1)}{4n^2+1}$$ The goal now is to find \(N\) such that the absolute value of the above expression is less than \(\epsilon\) for all \(n > N\).
02

Simplify the expression

Simplify the expression in the absolute value as follows: $$\frac{3n^2 - (3/4)(4n^2+1)}{4n^2+1} = \frac{3n^2 - 3n^2 - 3/4}{4n^2+1} = \frac{-3/4}{4n^2+1}$$
03

Handle the absolute value

Take the absolute value of the expression: $$\left|\frac{-3/4}{4n^2+1}\right| = \frac{3/4}{4n^2+1}$$
04

Bound the expression

We want to find an \(N\) such that for all \(n > N\) $$\frac{3/4}{4n^2+1} < \epsilon$$ Notice that \(4n^2+1 > 4n^2\) for all \(n\). Therefore, we can write $$\frac{3/4}{4n^2+1} < \frac{3/4}{4n^2}$$ Now, we want to find an \(N\) such that for all \(n > N\), $$\frac{3/4}{4n^2} < \epsilon$$
05

Find the value of N

We can now solve for \(N\) as follows: $$\frac{3/4}{4n^2} < \epsilon \Rightarrow n^2 > \frac{3/4}{4\epsilon} \Rightarrow n > \sqrt{\frac{3}{16\epsilon}}$$ Let $$N = \left\lceil\sqrt{\frac{3}{16\epsilon}}\right\rceil$$ where \(\lceil x \rceil\) denotes the smallest integer greater than or equal to \(x\).
06

Conclude the proof

Finally, we can conclude the proof by saying that for any \(\epsilon > 0\), we have found a positive integer \(N\) such that for all \(n > N\), we have $$\left|\frac{3 n^{2}}{4 n^{2}+1} - \frac{3}{4}\right| < \epsilon$$ Hence, using the formal definition of the limit, we have proved that $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$

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Most popular questions from this chapter

The sequence \(\\{n !\\}\) ultimately grows faster than the sequence \(\left\\{b^{n}\right\\},\) for any \(b>1,\) as \(n \rightarrow \infty .\) However, \(b^{n}\) is generally greater than \(n !\) for small values of \(n\). Use a calculator to determine the smallest value of \(n\) such that \(n !>b^{n}\) for each of the cases \(b=2, b=e,\) and \(b=10\).

A ball is thrown upward to a height of \(h_{0}\) meters. After each bounce, the ball rebounds to a fraction r of its previous height. Let \(h_{n}\) be the height after the nth bounce and let \(S_{n}\) be the total distance the ball has traveled at the moment of the nth bounce. a. Find the first four terms of the sequence \(\left\\{S_{n}\right\\}\) b. Make a table of 20 terms of the sequence \(\left\\{S_{n}\right\\}\) and determine \(a\) plausible value for the limit of \(\left\\{S_{n}\right\\}\) $$h_{0}=20, r=0.75$$

The famous Fibonacci sequence was proposed by Leonardo Pisano, also known as Fibonacci, in about \(\mathrm{A.D.} 1200\) as a model for the growth of rabbit populations. It is given by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1 .\) Each term of the sequence is the sum of its two predecessors. a. Write out the first ten terms of the sequence. b. Is the sequence bounded? c. Estimate or determine \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}},\) the ratio of the successive terms of the sequence. Provide evidence that \(\varphi=(1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Use induction to verify the remarkable result that $$f_{n}=\frac{1}{\sqrt{5}}\left(\varphi^{n}-(-1)^{n} \varphi^{-n}\right).$$

The Fibonacci sequence \(\\{1,1,2,3,5,8,13, \ldots\\}\) is generated by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1\). a. It can be shown that the sequence of ratios of successive terms of the sequence \(\left\\{\frac{f_{n+1}}{f_{n}}\right\\}\) has a limit \(\varphi .\) Divide both sides of the recurrence relation by \(f_{n},\) take the limit as \(n \rightarrow \infty,\) and show that \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}}=\frac{1+\sqrt{5}}{2} \approx 1.618\). b. Show that \(\lim _{n \rightarrow \infty} \frac{f_{n-1}}{f_{n+1}}=1-\frac{1}{\varphi} \approx 0.382\). c. Now consider the harmonic series and group terms as follows: $$\sum_{k=1}^{\infty} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)$$ $$+\left(\frac{1}{9}+\cdots+\frac{1}{13}\right)+\cdots$$ With the Fibonacci sequence in mind, show that $$\sum_{k=1}^{\infty} \frac{1}{k} \geq 1+\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{3}{8}+\frac{5}{13}+\cdots=1+\sum_{k=1}^{\infty} \frac{f_{k-1}}{f_{k+1}}.$$ d. Use part (b) to conclude that the harmonic series diverges. (Source: The College Mathematics Journal, 43, May 2012)

Evaluate the limit of the following sequences or state that the limit does not exist. $$a_{n}=\frac{75^{n-1}}{99^{n}}+\frac{5^{n} \sin n}{8^{n}}$$
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