# Proof that √x does not tend to a limit as x approaches infinity

I am wanting to prove from the definition of a limit ( $∀ε>0 ∃K>0:∀x>K, |f(x)−l|$<ε) that $√x$ does not tend to a limit as $x$ approaches infinity.

So far I have tried to find a value of K such that $|f(x)−l|$<ε holds, and was planing on picking an $x$ bigger than K that would give $f(x)>l$, however I am having trouble finding a value for K.

Thanks for any suggestions

• You just want to show that the square root is unbounded. So, for any $N$ you can find $x$ such that $\sqrt{x}\gt N$. $x=(N+1)^2$ should do the trick. – John Douma Nov 6 '15 at 21:46
• But how would I then prove that it doesn't tend to a limit from this? – user284408 Nov 6 '15 at 21:58
• If it tends to a limit that limit is a real number and so it is bounded. – John Douma Nov 6 '15 at 21:59
• You can use $K=(\epsilon +l)^2$. – Emilio Novati Nov 6 '15 at 22:17

## 1 Answer

By definition, $\lim_{x\to\infty}f(x)=L$ if $\forall\epsilon\gt 0$ there exists $N\gt 0$ such that $x\gt N\implies |f(x)-L|\lt\epsilon$

$|f(x)-L|\lt\epsilon$ is equivalent to $L-\epsilon\lt f(x)\lt L+\epsilon$.

To show that the limit does not exist, we can show that for any $L$, there is a fixed $\epsilon$ for which the inequality cannot hold.

Since we are talking about the square root we can assume $L$ is positive. Let $\epsilon = 1$. Then it is sufficient to show that we can always find $x$ such that $\sqrt{x}\ge L+1$.

Let $x=max(N+1, (L+2)^2)$. Then $x\gt N$ and $\sqrt{x}=\sqrt{(L+2)^2}=L+2\gt L+1$ so $L$ cannot be the limit. Since this is true for all $L$, the square root does not converge to a limit as $x$ approaches infinity.

• If $a$ is not less than $b$ then $a$ is greater than or equal to $b$. Is that the confusion or are you asking why $\sqrt{x}\ge L+1$? – John Douma Nov 6 '15 at 22:34
• I understand now thankyou! I thought you were rearranging the top inequality to that which is the complete opposite – user284408 Nov 6 '15 at 22:36