Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
To determine the limit of the sequence as [tex]\( n \)[/tex] approaches infinity, we start by analyzing the given expression for [tex]\( a_n \)[/tex]:
[tex]\[ a_n = \frac{4 + n - 4n^2}{5n^2 + 9} \][/tex]
Step 1: Identify the highest power of [tex]\( n \)[/tex] in both the numerator and the denominator. In this case, the highest power of [tex]\( n \)[/tex] in the numerator is [tex]\( n^2 \)[/tex] and in the denominator is also [tex]\( n^2 \)[/tex].
Step 2: To simplify the expression, we divide every term in the numerator and the denominator by [tex]\( n^2 \)[/tex]:
[tex]\[ a_n = \frac{\frac{4}{n^2} + \frac{n}{n^2} - 4}{5 + \frac{9}{n^2}} \][/tex]
[tex]\[ a_n = \frac{\frac{4}{n^2} + \frac{1}{n} - 4}{5 + \frac{9}{n^2}} \][/tex]
Step 3: As [tex]\( n \)[/tex] approaches infinity, the terms [tex]\(\frac{4}{n^2}\)[/tex], [tex]\(\frac{1}{n}\)[/tex], and [tex]\(\frac{9}{n^2}\)[/tex] all approach 0. Thus, the expression simplifies to:
[tex]\[ a_n = \frac{0 + 0 - 4}{5 + 0} = \frac{-4}{5} = -\frac{4}{5} \][/tex]
Therefore, the limit of the sequence as [tex]\( n \)[/tex] approaches infinity is:
[tex]\[ \lim_ {n \rightarrow \infty} a_n = -\frac{4}{5} \][/tex]
[tex]\[ a_n = \frac{4 + n - 4n^2}{5n^2 + 9} \][/tex]
Step 1: Identify the highest power of [tex]\( n \)[/tex] in both the numerator and the denominator. In this case, the highest power of [tex]\( n \)[/tex] in the numerator is [tex]\( n^2 \)[/tex] and in the denominator is also [tex]\( n^2 \)[/tex].
Step 2: To simplify the expression, we divide every term in the numerator and the denominator by [tex]\( n^2 \)[/tex]:
[tex]\[ a_n = \frac{\frac{4}{n^2} + \frac{n}{n^2} - 4}{5 + \frac{9}{n^2}} \][/tex]
[tex]\[ a_n = \frac{\frac{4}{n^2} + \frac{1}{n} - 4}{5 + \frac{9}{n^2}} \][/tex]
Step 3: As [tex]\( n \)[/tex] approaches infinity, the terms [tex]\(\frac{4}{n^2}\)[/tex], [tex]\(\frac{1}{n}\)[/tex], and [tex]\(\frac{9}{n^2}\)[/tex] all approach 0. Thus, the expression simplifies to:
[tex]\[ a_n = \frac{0 + 0 - 4}{5 + 0} = \frac{-4}{5} = -\frac{4}{5} \][/tex]
Therefore, the limit of the sequence as [tex]\( n \)[/tex] approaches infinity is:
[tex]\[ \lim_ {n \rightarrow \infty} a_n = -\frac{4}{5} \][/tex]
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.