Why the Limit of $\sqrt x$ is infinity?

Question

I'm a college student taking calculus for the first time. This past two weeks we were learning about limits and continuity. We are using Calculus Early Transcendentals book by Anton, Bivens, and Davis. Question 11 from exercise set 1.3 asks to find $\lim_{x\to \infty} \sqrt{x}$. The book answer for $\lim_{x\to \infty} \sqrt{x}$ is $+\infty$. Can anyone please explain me why the result is positive infinity. I do not know the procedure the book used to get the answer.

Answer 1

This is the graph of $y=\sqrt{x}$.

You can see clearly that as $x\to\infty \implies y\to \infty$ as well.

Answer 2

If you take the square root of numbers that grow without bound, the result grows without bound.

Answer 3

$N$	$\sqrt{N}$ (approx.)
$5$	$2.236$
$50$	$7.071$
$2000$	$44.72$
$\vdots$	$\vdots$

Keep taking larger and larger $N$, and you will see, while $N$ increases, $\sqrt{N}$ also increases.
Further, this is an unbounded increase! That is, you can make $\sqrt{N}$ as large as possible by taking a large enough $N$. That is, $\sqrt{N}$ can be as large as it wants - there is no limit - that's why the limit is infinite!

Answer 4

If you take the square of an arbitarily large number, the result is also large but always finite. Therefore, it's always possible to find a large enough $x$ where $\sqrt{x}$ is arbitrarily large, and so the limit is infinity. $$\sqrt{1}=1$$ $$\sqrt{100}=10$$ $$\sqrt{1000000}=1000$$ etc.