find the limit of the following sequence or determine that the sequence diverges. {n / √(16n² + 7)} select…

Explanation:

Step1: Divide numerator and denominator by n

Divide both the numerator and denominator of $\frac{n}{\sqrt{16n^{2}+7}}$ by $n$. Since $n=\sqrt{n^{2}}$ for $n > 0$, we get $\lim_{n\rightarrow\infty}\frac{n}{\sqrt{16n^{2}+7}}=\lim_{n\rightarrow\infty}\frac{n/n}{\sqrt{16n^{2}+7}/n}=\lim_{n\rightarrow\infty}\frac{1}{\sqrt{\frac{16n^{2}+7}{n^{2}}}}$.

Step2: Simplify the denominator

Simplify $\sqrt{\frac{16n^{2}+7}{n^{2}}}=\sqrt{16 + \frac{7}{n^{2}}}$. So the limit becomes $\lim_{n\rightarrow\infty}\frac{1}{\sqrt{16+\frac{7}{n^{2}}}}$.

Step3: Evaluate the limit

As $n\rightarrow\infty$, $\frac{7}{n^{2}}\rightarrow0$. Then $\lim_{n\rightarrow\infty}\frac{1}{\sqrt{16+\frac{7}{n^{2}}}}=\frac{1}{\sqrt{16 + 0}}$.

Step4: Calculate the final value

$\frac{1}{\sqrt{16+0}}=\frac{1}{4}$.

Answer:

A. The limit of the sequence is $\frac{1}{4}$