How does this limit with a square root behave as it approaches infinity? $$
\lim_{x\to \infty}\frac{2x}{\sqrt{4x^2+1}}
$$
Can I substitute $\sqrt{4x^2}$ for $2x$ in the denominator to show that this limit approaches $1$?

For fun:
Let $x >0.$
$\dfrac{2x}{\sqrt{4x^2 +8x +1}} \lt \dfrac{2x}{\sqrt{4x^2+1} }\lt $
$\dfrac{2x}{\sqrt{4x^2}}$.
$\dfrac{2x}{\sqrt{(2x+1)^2}}\lt \dfrac{2x}{\sqrt{4x^2+1}} \lt $
$\dfrac{2x}{\sqrt{4x^2}}.$
Note : $\sqrt{x^2} = |x| =x,$ since $x>0.$
$\dfrac{2x}{\sqrt{(2x+1)^2}} = \dfrac{2x}{2x+1}= 
\dfrac{2}{2+1/x} $.
We have:
$\dfrac {2}{2+1/x} \lt \dfrac {2x}{\sqrt{4x^2 +1}} \lt \dfrac{2x}{2x}=1$.
The limit is?
