Find Delta, given Epsilon lim x->0 root (x+1) = 1. epsilon = 0.1 Can I solve this using rationalization and then assuming delta <= 1? If so, why is my answer wrong?
we know |√(x+1)-1| < 0.1
rationalization of LHS,
x<0.1*(|√(x+1)+1|)
hence, delta = 0.1*(|√(x+1)+1|)
to find x, consider delta <= 1, x <= 1 hence √(x+1)+1 <= √2 + 1
substiting the value, delta = 0.1*(√2 + 1) = 0.2 (incorrect answer)


There are two mistakes you are making:


*

*The $\delta$ you are choosing in the first place is not independent of $x$.

*In trying to rectify this, you find an upper bound $c$ for your expression $f(x)$ for $\delta$ that is dependent on $x$ and then choose $\delta$ equal to that upper bound $c$. However, you previously found that $\delta$ must be less than or equal to $f(x)$ for all $x$. But you have proven your chosen $\delta=c$ to be greater than $f(x)$ for $x<1$. In conclusion, this approach cannot work.


Rather, consider the following: We have, as you already noted
$$x<\varepsilon\cdot \left|\sqrt{x+1}+1\right|.$$
Since the expression inside the absolute value is always positive, we can drop the absolute value and obtain
$$\begin{split}
x&<\varepsilon\cdot\left(\sqrt{x+1}+1\right) \\
\Leftrightarrow x-\varepsilon&<\varepsilon\cdot\sqrt{x+1} \\
\Leftrightarrow x^2-2\varepsilon x+\varepsilon^2&<\varepsilon x+\varepsilon \\
\Leftrightarrow x^2-3\varepsilon x+(\varepsilon^2-\varepsilon)&<0 \\
\end{split}$$
We will find the zeroes of the quadratic function on the left. Since its leading term is positive, all $x$ between those zeroes will satisfy the inequality.
$$x_{1,2}=\frac{3}{2}\varepsilon\pm\sqrt{\frac{9}{4}\varepsilon^2-\varepsilon^2+\varepsilon}=\frac{3}{2}\varepsilon\pm\sqrt{\frac{5}{4}\varepsilon^2+\varepsilon}.$$
In the case of $\varepsilon=0.1$, we get $x_1=\frac{3+3\sqrt{5}}{20}$ and $x_2=\frac{3-3\sqrt{5}}{20}$. We have $|x_1|>|x_2|$ and thus we have to take $\delta=|x_2|=\frac{3\sqrt{5}-3}{20}$.
