Can a Limit Converging to the Square Root of x be Proven from Given Statements?

[3102]

Homework Statement

I have given the statements: $a_{n}^2 \ge x$ , $a_{n+1} \le a_{n}$ , $x > 0$ and $\inf a_{n} > 0$. How to prove the following: $\lim_{n \to \infty}a_{n}=\sqrt{x}$

Homework Equations

$a_{n}^2 \ge x$ , $a_{n+1} \le a_{n}$ , $x > 0$ and $\inf a_{n} > 0$
$\lim_{n \to \infty}a_{n}=\sqrt{x}$

The Attempt at a Solution

I have come so far: $$a_{n}\ge a_{n+1} \ge \sqrt{x}$$ How shall I continue?

 

[PeroK]

Homework Statement

I have given the statements: $a_{n}^2 \ge x$ , $a_{n+1} \le a_{n}$ , $x > 0$ and $\inf a_{n} > 0$. How to prove the following: $\lim_{n \to \infty}a_{n}=\sqrt{x}$

Homework Equations

$a_{n}^2 \ge x$ , $a_{n+1} \le a_{n}$ , $x > 0$ and $\inf a_{n} > 0$
$\lim_{n \to \infty}a_{n}=\sqrt{x}$

The Attempt at a Solution

I have come so far: $$a_{n}\ge a_{n+1} \ge \sqrt{x}$$ How shall I continue?

Are you sure that's the problem? There's not enough information there to show that the limit is $\sqrt{x}$

 

[HallsofIvy]

You can't prove this, it is not true!

For example, a decreasing sequence that converges to $\sqrt{x+ 1}$ will satisfy all the hypotheses of your "theorem".