Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

Determine the limit of the sequence or state that the sequence diverges.

[tex]a_n = \frac{4+n-4n^2}{5n^2+9}[/tex]

(Use symbolic notation and fractions where needed. Enter DNE if the sequence diverges.)

[tex]\lim_{n \rightarrow \infty} a_n = \square[/tex]

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]