I understand that I'm supposed to manipulate the expression of $\lvert f(x) -L \rvert$, i.e. $\left\lvert \sqrt{x^2 – x + 3} – 3 \right\rvert$ to somehow extricate a $\lvert x- a\rvert$, i.e. $\lvert x-3 \rvert$ so that I can express $\delta$ in terms of $\varepsilon$. Here's what I've done so far, and where I've gotten stuck.
Given $\varepsilon \gt 0$, choose $\delta = \_\_\_\_$ (TBD).
Then, for $0 \lt \lvert x – 3 \rvert \lt \delta$,
\begin{align}
\lvert f(x) -L \rvert &= \left\lvert \sqrt{x^2 – x + 3} – 3 \right\rvert \\
& = \left\lvert \sqrt{x^2 – x + 3} – 3 \; \cdot \; \frac{\sqrt{x^2 – x + 3} +3}{\sqrt{x^2 – x + 3} + 3}\right\rvert \\
& = \left\lvert \frac{x^2 – x – 6}{\sqrt{x^2 – x + 3} + 3 } \right\rvert \\
& = \left\lvert\frac{(x – 3)(x+2)}{\sqrt{x^2 – x + 3} + 3 }\right\rvert \\
& \lt \frac{\delta(x – 3 + 5)}{\sqrt{x^2 – x + 3} + 3} \\
& \lt \frac{\delta(\delta + 5)}{\sqrt{x^2 – x + 3} + 3} \\
& = \; ???
\end{align}
What do I do with the denominator? Or is my method incorrect from the get-go?
Best Answer
Notice that when $\delta<1$,
$$|\frac{\delta(x+2)}{\sqrt{x^2-x+3}+3}|\leq \delta|x+2|/3,$$
As the denominator is at least 3 (really we're just ensuring the square root is real). Now notice that $|x+2|\leq (|x-3|+|5|)\leq\delta+5$. So now pick $\delta$ to be small enough.