Let's determine the limit of the sequence [tex]\( a_n = 6 - \frac{3}{n^2} \)[/tex] as [tex]\( n \)[/tex] approaches infinity.
1. We start by examining the given sequence:
[tex]\[
a_n = 6 - \frac{3}{n^2}
\][/tex]
2. To find the limit as [tex]\( n \)[/tex] approaches infinity, we analyze each term separately:
- The term [tex]\( 6 \)[/tex] remains constant as [tex]\( n \)[/tex] approaches infinity.
- The term [tex]\( \frac{3}{n^2} \)[/tex] changes as [tex]\( n \)[/tex] grows larger. Since [tex]\( n^2 \)[/tex] becomes very large as [tex]\( n \)[/tex] approaches infinity, [tex]\( \frac{3}{n^2} \)[/tex] becomes very small.
3. Formally, as [tex]\( n \)[/tex] approaches infinity, [tex]\( n^2 \rightarrow \infty \)[/tex].
4. Hence, [tex]\( \frac{3}{n^2} \rightarrow 0 \)[/tex].
5. Therefore, the sequence [tex]\( 6 - \frac{3}{n^2} \)[/tex] approaches:
[tex]\[
6 - 0 = 6.
\][/tex]
6. Thus, we conclude that the limit of the sequence [tex]\( a_n \)[/tex] as [tex]\( n \)[/tex] approaches infinity is:
[tex]\[
\lim_{n \rightarrow \infty} a_n = 6
\][/tex]
So, the limit of [tex]\( a_n \)[/tex] is [tex]\(\boxed{6}\)[/tex]. The initial value given in the problem ([tex]\( \lim_{n \rightarrow \infty} a_n = 1 \)[/tex]) appears to be incorrect based on the analysis.