The limit of the sequence $a _ { n } = \frac { 3 \sqrt { n } + 2 } { \sqrt { n } + 5 }$ is

A)0
B)3
C) $\sqrt { 3 }$
D) $\frac { 2 } { 5 }$
E)$\infty$

Question

The limit of the sequence $a _ { n } = \frac { 3 \sqrt { n } + 2 } { \sqrt { n } + 5 }$ is

A)0
B)3
C) $\sqrt { 3 }$
D) $\frac { 2 } { 5 }$
E)$\infty$

Quizplus · Accepted Answer

To find the limit of the sequence as $n$ approaches infinity, divide the numerator and the denominator by $\sqrt{n}$. This simplifies the expression to $\frac{3 + \frac{2}{\sqrt{n}}}{1 + \frac{5}{\sqrt{n}}}$. As $n$ approaches infinity, the terms $\frac{2}{\sqrt{n}}$ and $\frac{5}{\sqrt{n}}$ approach 0, leaving the limit as $\frac{3}{1} = 3$.

The Limit of the Sequence an=3n+2n+5a _ { n } = \frac { 3 \sqrt { n } + 2 } { \sqrt { n } + 5 }an​=n​+53n​+2​

The Limit of the Sequence $a _ { n } = \frac { 3 \sqrt { n } + 2 } { \sqrt { n } + 5 }$