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Chapter 7: Problem 4

Evaluate \(\lim _{n \rightarrow \infty} \frac{7 n^{2}-4 n+3}{3 n^{2}+5 n+9}\)

Short Answer

Expert verified
The limit of the given expression as n approaches infinity is \(\frac{7}{3}\).

Step by step solution

01

Identify the degree of the polynomials

Start by identifying the degree of both the numerator and the denominator of the given expression. The degree of a polynomial is the highest power of the variable in the polynomial. In this case: For the numerator: \(7n^2 - 4n + 3\), the degree is 2, as the highest power of n is 2. For the denominator: \(3n^2 + 5n + 9\), the degree is also 2, for the same reason. Since both polynomials have the same degree, we can find the limit by dividing the leading coefficients.
02

Divide the leading coefficients

The leading coefficients are the coefficients of the highest degree terms in both polynomial. For the numerator, the leading coefficient is 7 (from the term \(7n^2\)), and for the denominator, the leading coefficient is 3 (from the term \(3n^2\)). Now, we divide the leading coefficients 7 and 3: \(\lim_{n \rightarrow \infty} \frac{7n^2 - 4n + 3}{3n^2 + 5n + 9} = \frac{7}{3}\) The limit of the given expression as n approaches infinity is \(\frac{7}{3}\).

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Most popular questions from this chapter

In Exercises \(1-10,\) evaluate the arithmetic series. \(\sum_{k=5}^{65}(4 k-1)\)

Consider the sequence whose \(n^{\text {th }}\) term \(a_{n}\) is given by the indicated formula. (a) Write the sequence using the three-dot notation, giving the first four terms of the sequence. (b) Write the sequence as a recursive sequence. \(a_{n}=5 \cdot 3^{-n}\)

Explain why 0.2 and the repeating decimal \(0.199999 \ldots\) both represent the real number \(\frac{1}{5}\).

Give the first four terms of the specified recursive sequence. \(a_{1}=2\) and \(a_{n+1}=3 a_{n}-5\) for \(n \geq 1\)

Evaluate \(\lim _{n \rightarrow \infty} n\left(e^{1 / n}-1\right)\).
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