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Chapter 7: Problem 3

Evaluate \(\lim _{n \rightarrow \infty} \frac{2 n^{2}+5 n+1}{5 n^{2}-6 n+3}\)

Short Answer

Expert verified
The limit of the given function \(\frac{2n^2 + 5n + 1}{5n^2 - 6n + 3}\) as n approaches infinity is \(\frac{2}{5}\).

Step by step solution

01

Identify the dominant terms

The exercise asks us to find the limit of the function, so we must analyze the terms in the given rational function as n approaches infinity. Given function: \(\frac{2n^2 + 5n + 1}{5n^2 - 6n + 3}\) For large n, the dominant terms are those with the highest power of n. In this case, the dominant terms are \(2n^2\) in the numerator and \(5n^2\) in the denominator.
02

Simplify the function

As we now know that the dominant terms are \(2n^2\) in the numerator and \(5n^2\) in the denominator, let's divide both the numerator and the denominator by \(n^2\): Simplified function: \(\frac{\dfrac{2n^2}{n^2} + \dfrac{5n}{n^2} + \dfrac{1}{n^2}}{\dfrac{5n^2}{n^2} - \dfrac{6n}{n^2} + \dfrac{3}{n^2}}\) Simplifying further, we get: Simplified function: \(\frac{2 + \dfrac{5}{n} + \dfrac{1}{n^2}}{5 - \dfrac{6}{n} + \dfrac{3}{n^2}}\)
03

Evaluate the limit

Now we'll evaluate the limit as n approaches infinity: \(\lim _{n\rightarrow\infty} \frac{2 + \dfrac{5}{n} + \dfrac{1}{n^2}}{5 - \dfrac{6}{n} + \dfrac{3}{n^2}}\) As n approaches infinity, \(n\) will become very large, causing the terms \(\dfrac{5}{n}\), \(\dfrac{1}{n^2}\), \(\dfrac{6}{n}\), and \(\dfrac{3}{n^2}\) to approach zero since their denominators are getting larger. Thus, we have: \(\lim _{n\rightarrow\infty} \frac{2 + \dfrac{5}{n} + \dfrac{1}{n^2}}{5 - \dfrac{6}{n} + \dfrac{3}{n^2}} = \frac{2}{5}\) So, the limit of the given function as n approaches infinity is \(\frac{2}{5}\).

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