Proof that $\lim _{x\to \infty }(\sqrt{x^2+2x}-x)=1$ using the epsilon delta definition of limit

Question

Starting with, $$|\sqrt{x^2+2x}-x-1|<\epsilon$$

So my initial attempt to this problem is to factor out the $x$ out: $$|x||\sqrt{1+\frac{2}{x}} -x^2-\frac{1}{x}|<\epsilon$$

and then we can restrict $x>2$, such that $\sqrt{1+\frac{2}{x}} -x^2-\frac{1}{x}<\sqrt2-1-\frac{1}{2}$

so, $$|x|<\frac{\epsilon}{\sqrt2-1-\frac{1}{2}}$$

However, I just realized that my target was supposed to be $|x|>M$, not $|x|<M$, so I cannot proceed with the final inequality I found above. I tried factoring out $|1/x|$ for the initial step with the hope of getting to $|\frac{1}{x}|<M$, but I am stuck.

So how can I find the suitable $M$ needed for this problem?

Thank you

Answer 1

Hint: Let $x > 0$ to begin with. $\left|\sqrt{x^2+2x} - x-1\right|= \dfrac{1}{\sqrt{x^2+2x} + x+1}< \dfrac{1}{x+1}< \epsilon$. Can you solve for $x$ in the last inequality and then find $M$ ?