- #1

- 10

- 0

Just hoping someone could check my work and point out any errors, if any.

## Homework Statement

Consider the sequence {[itex]a_n[/itex]} defined by [itex]a_n=\frac{n}{2n+\sqrt{n}}[/itex]. Prove that [itex]\lim_{x\to\infty}a_n=\frac{1}{2}[/itex]. (Do NOT use any of the "limit rules" from Section 2.2.)

## Homework Equations

A sequence [itex]a_n[/itex] is said to converge to a real number A iff for each [itex]\epsilon>0[/itex] there exists a positive integer [itex]n^*[/itex] such that [itex]\lvert a_n-A\rvert<\epsilon[/itex] for all [itex]n\geq n^*[/itex].

## The Attempt at a Solution

**Proof:**

Let [itex]\epsilon>0[/itex].

Let [itex]n^*\in\mathbb{N}[/itex] such that [itex]n^*>[\frac{1}{2}(\frac{1}{\epsilon}-1)]^2[/itex].

If [itex]n\geq n^*[/itex], then

[itex]\lvert a_n-A\rvert[/itex]

[itex]=\lvert\frac{n}{2n+\sqrt{n}}-\frac{1}{2}\rvert[/itex]

[itex]=\lvert\frac{2n-2n-\sqrt{n}}{2(2n+\sqrt{n})}\rvert[/itex]

[itex]=\lvert\frac{-\sqrt{n}}{2(2n+\sqrt{n})}\rvert[/itex]

[itex]=\lvert\frac{-n}{\sqrt{n}[2(2n+\sqrt{n})]}\rvert[/itex]

[itex]=\lvert\frac{-n}{\sqrt{n}(4n+2\sqrt{n})}\rvert[/itex]

[itex]=\lvert\frac{-n}{4n\sqrt{n}+2n}\rvert[/itex]

[itex]=\lvert\frac{-n}{2n(2\sqrt{n}+1)}\rvert[/itex]

[itex]=\lvert\frac{-1}{2(2\sqrt{n}+1)}\rvert[/itex]

[itex]=\frac{1}{2(2\sqrt{n}+1)}[/itex]

[itex]\leq\frac{1}{2\sqrt{n}+1}[/itex]

[itex]<\frac{1}{2\sqrt{[\frac{1}{2}(\frac{1}{\epsilon}-1)]^2}+1}[/itex]

[itex]=\frac{1}{2\lvert[\frac{1}{2}(\frac{1}{\epsilon}-1)]\rvert+1}[/itex]

[itex]=\frac{1}{2[\frac{1}{2}(\frac{1}{\epsilon}-1)]+1}[/itex]

[itex]=\frac{1}{(\frac{1}{\epsilon}-1)+1}[/itex]

[itex]=\frac{1}{\frac{1}{\epsilon}}[/itex]

[itex]=\epsilon[/itex].