To determine the limit of the sequence given by [tex]\( a_n = e^{\frac{2n}{3n + 6}} \)[/tex] as [tex]\( n \)[/tex] approaches infinity, follow these steps:
1. Analyze the exponent: We first focus on the expression inside the exponent, which is [tex]\( \frac{2n}{3n + 6} \)[/tex].
2. Simplify the fraction: To simplify [tex]\( \frac{2n}{3n + 6} \)[/tex], divide both the numerator and the denominator by [tex]\( n \)[/tex]:
[tex]\[
\frac{2n}{3n + 6} = \frac{2n / n}{(3n + 6) / n} = \frac{2}{3 + \frac{6}{n}}
\][/tex]
3. Evaluate the limit inside the exponent: As [tex]\( n \)[/tex] approaches infinity, the term [tex]\( \frac{6}{n} \)[/tex] approaches 0. Therefore, the expression [tex]\( 3 + \frac{6}{n} \)[/tex] approaches 3, and so:
[tex]\[
\lim_{n \to \infty} \frac{2}{3 + \frac{6}{n}} = \frac{2}{3}
\][/tex]
4. Apply the limit to the entire expression: Now we evaluate the limit of the entire expression [tex]\( a_n \)[/tex]. Since the exponent [tex]\( \frac{2n}{3n + 6} \)[/tex] approaches [tex]\( \frac{2}{3} \)[/tex] as [tex]\( n \)[/tex] approaches infinity, we get:
[tex]\[
\lim_{n \to \infty} a_n = \lim_{n \to \infty} e^{\frac{2n}{3n + 6}} = e^{\frac{2}{3}}
\][/tex]
Therefore, the limit of the sequence [tex]\( a_n = e^{\frac{2n}{3n + 6}} \)[/tex] as [tex]\( n \)[/tex] approaches infinity is:
[tex]\[
\lim_{n \to \infty} a_n = e^{\frac{2}{3}}
\][/tex]
