# Prove that $\lim_{x \to 3}{\sqrt{x^2 - x + 3}} =3$ using the precise definition of limits...

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?

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

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.