Question

a_n = \frac{9n}{\sqrt{49n^{2}+1}}
\lim_{n\to\infty} a_n =

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 $\lim_{n
ightarrow\infty}\frac{9n}{\sqrt{49n^{2}+1}}=\lim_{n
ightarrow\infty}\frac{9}{\sqrt{49 + \frac{1}{n^{2}}}}$.

Step2: Evaluate the limit

As $n
ightarrow\infty$, $\frac{1}{n^{2}}
ightarrow0$. So $\lim_{n
ightarrow\infty}\frac{9}{\sqrt{49+\frac{1}{n^{2}}}}=\frac{9}{\sqrt{49 + 0}}$.

Step3: Simplify the expression

$\frac{9}{\sqrt{49+0}}=\frac{9}{7}$.

Answer:

$\frac{9}{7}$