a_n = \\frac{9n}{\\sqrt{49n^{2}+1}}\n\\lim_{n\\to\\infty} a_n =
a_n = \\frac{9n}{\\sqrt{49n^{2}+1}}\n\\lim_{n\\to\\infty} a_n =
Answer
Explanation:
Step1: Divide numerator and denominator by $n$
Divide both the numerator $9n$ and the denominator $\sqrt{49n^{2}+1}$ by $n$. Since $n=\sqrt{n^{2}}$ for $n>0$, we have $\lim_{n\rightarrow\infty}\frac{9n}{\sqrt{49n^{2}+1}}=\lim_{n\rightarrow\infty}\frac{9}{\sqrt{49 + \frac{1}{n^{2}}}}$.
Step2: Evaluate the limit
As $n\rightarrow\infty$, $\frac{1}{n^{2}}\rightarrow0$. So $\lim_{n\rightarrow\infty}\frac{9}{\sqrt{49+\frac{1}{n^{2}}}}=\frac{9}{\sqrt{49 + 0}}$.
Step3: Simplify the expression
$\frac{9}{\sqrt{49+0}}=\frac{9}{7}$.
Answer:
$\frac{9}{7}$