Finishing my proof by $\epsilon$-$\delta$ definition that $\lim_{x \to 1} \sqrt{2x-1}-x =0$

Question

so far I have this in my proof:
Let $\epsilon >0$ so $|\sqrt{2x-1}-x|<\epsilon$ then: $$|\sqrt{2x-1}-x|=|\sqrt{2x-1}-x||\dfrac{\sqrt{2x-1}+x}{\sqrt{2x-1}+x}|=|\dfrac{2x-1-x^2}{\sqrt{2x-1}+x}|=|\dfrac{(x-1)^2}{\sqrt{2x-1}+x}|<\epsilon $$ Taking square root both sides: $$ |\dfrac{(x-1)}{\sqrt{\sqrt{2x-1}+x}}|<\sqrt{\epsilon} $$ I feel that I'm near the solution, 'cause I have the $|(x-1)|$ bounded by a function of $\epsilon$ but I don't know what to do with the denominator, somebody told me that I can just erase it and $\delta_{\epsilon}=\sqrt{\epsilon}$ but that would work iff the denominator is less than 1.

Answer 1

Hint:

Massage your expression into the form

$$\left| \frac{(x-1)^2}{\sqrt{2x-1} + x} \right| < M |x-1|$$ for some appropriate choice of $M > 0$. Do that by assuming for instance $|x-1| < \frac 14$, i.e., $\frac 34 < x < \frac 54$ and

$$\left| \frac{(x-1)^2}{\sqrt{2x-1} + x} \right| = \left| \frac{x-1}{\sqrt{2x-1} + x} \right| \left| x - 1 \right| < \frac 14 \left| \frac{1}{\sqrt{2x-1} + x} \right| \left| x - 1 \right| < \cdots $$

Having done that you finish this way: given $\epsilon > 0$, choose $\delta = \min\left( \frac 14, \frac \epsilon M\right)$.

Answer 2

$\text{Observe that if you take} :|x-1| < \dfrac{1}{3} \Rightarrow -\dfrac{1}{3} < x - 1 < \dfrac{1}{3} \Rightarrow \dfrac{2}{3} < x < \dfrac{4}{3} \Rightarrow \sqrt{2x-1} + x > x > \dfrac{2}{3} \Rightarrow \dfrac{|x-1|^2}{\sqrt{2x-1}+x} < \dfrac{|x-1|^2}{\dfrac{2}{3}}=\dfrac{3|x-1|^2}{2}< \epsilon \iff |x-1| < \sqrt{\dfrac{2\epsilon}{3}}\Rightarrow \text{ if we take $\delta = \text{min} \left(\dfrac{1}{3}, \sqrt{\dfrac{2\epsilon}{3}}\right)$} \text{ for a given $\epsilon > 0$, then we're done}.$