Question

1. -/1.11 pointsdetermine whether the sequence converges or diverges. if it converges, find the limit. (if the sequence diverges, enter diverges.)( a _ { n } = e ^ { - 7 / sqrt { n } } )( lim _ { n

ightarrow infty } a _ { n } = )8. -/1.11 pointsdetermine whether the sequence converges or diverges. if it converges, find the limit. (if the sequence diverges, enter diverges.)( a _ { n } = \frac { 3 ^ { n } } { 1 + 7 ^ { n } } )( lim _ { n
ightarrow infty } a _ { n } = )

Explanation:

Step1: Analyze the limit of the exponent for \(a_{n}=e^{-7/\sqrt{n}}\)

As \(n
ightarrow\infty\), \(\sqrt{n}
ightarrow\infty\), so \(\frac{7}{\sqrt{n}}
ightarrow0\). Then \(-\frac{7}{\sqrt{n}}
ightarrow0\).

Step2: Use the property of the exponential function

Since \(y = e^{x}\) is a continuous function, \(\lim_{n
ightarrow\infty}a_{n}=\lim_{n
ightarrow\infty}e^{-7/\sqrt{n}}\). By the continuity of \(y = e^{x}\), \(\lim_{n
ightarrow\infty}e^{-7/\sqrt{n}}=e^{\lim_{n
ightarrow\infty}(-7/\sqrt{n})}\). Substituting \(\lim_{n
ightarrow\infty}(-7/\sqrt{n}) = 0\), we get \(e^{0}\).

Step3: Calculate \(e^{0}\)

We know that for any real - number \(a\), \(a^{0}=1\) (\(a
eq0\)), so \(e^{0}=1\).

Step4: Analyze the limit for \(a_{n}=\frac{3^{n}}{1 + 7^{n}}\)

Divide both the numerator and denominator by \(7^{n}\). Then \(a_{n}=\frac{3^{n}/7^{n}}{1/7^{n}+7^{n}/7^{n}}=\frac{(\frac{3}{7})^{n}}{\frac{1}{7^{n}} + 1}\).

Step5: Use the limit property \(\lim_{n

ightarrow\infty}r^{n}=0\) for \(|r|\lt1\)
As \(n
ightarrow\infty\), since \(|\frac{3}{7}|\lt1\) and \(|\frac{1}{7}|\lt1\), \(\lim_{n
ightarrow\infty}(\frac{3}{7})^{n}=0\) and \(\lim_{n
ightarrow\infty}\frac{1}{7^{n}}=0\).

Step6: Calculate the limit of \(a_{n}=\frac{(\frac{3}{7})^{n}}{\frac{1}{7^{n}} + 1}\)

By the quotient rule of limits \(\lim_{n
ightarrow\infty}\frac{f(n)}{g(n)}=\frac{\lim_{n
ightarrow\infty}f(n)}{\lim_{n
ightarrow\infty}g(n)}\) (\(\lim_{n
ightarrow\infty}g(n)
eq0\)), \(\lim_{n
ightarrow\infty}\frac{(\frac{3}{7})^{n}}{\frac{1}{7^{n}}+1}=\frac{\lim_{n
ightarrow\infty}(\frac{3}{7})^{n}}{\lim_{n
ightarrow\infty}\frac{1}{7^{n}}+\lim_{n
ightarrow\infty}1}\). Substituting \(\lim_{n
ightarrow\infty}(\frac{3}{7})^{n}=0\) and \(\lim_{n
ightarrow\infty}\frac{1}{7^{n}}=0\), we get \(\frac{0}{0 + 1}=0\).

Answer:

For \(a_{n}=e^{-7/\sqrt{n}}\), the limit is \(1\).
For \(a_{n}=\frac{3^{n}}{1 + 7^{n}}\), the limit is \(0\).