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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?

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  • $\begingroup$ You could try to consider $\sqrt {x^2-x+3} > \Box$ where $\Box$ is a constant, then the fraction would be simpler. $\endgroup$
    – xbh
    Sep 28, 2018 at 3:05

1 Answer 1

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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.

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