Prove using epsilon - delta

let $f(x) = \dfrac {x-\sqrt x} {5x-4}$

prove using $\epsilon-\delta$ that $\lim_{x\to1} f(x) = 0$

I got it to -

let $\epsilon >0$.

choose $\delta = min(0.1 , \dfrac \epsilon 2)$

Than for all $\lvert x- 1\rvert < \delta$ :

$\lvert \dfrac {x-\sqrt x} {5x-4} \rvert $ = $\lvert \dfrac {x^2 - x} {(5x-4)(x+\sqrt x)} \rvert \le \dfrac {2\lvert x\rvert \lvert x -1\rvert} {\lvert x+\sqrt x\rvert}$$< 2\delta$ = $2\dfrac\epsilon 2$ = $\epsilon$

The part im not sure about is this :

$\dfrac {\lvert x\rvert } {\lvert x+\sqrt x\rvert} < 1$ for all $\lvert x- 1\rvert < \delta$

For some reason i think its too simple and that i might missed somthing here. would appreciate any clarifications.

Thanks.