# How do you use continuity to evaluate the limit arctan(x^2-4)/(3x^2-6x)?

Jun 24, 2015

If $x \to a$, where $a \ne 0$ and $a \ne 2$, then ${\lim}_{x \to a} \arctan \frac{{x}^{2} - 4}{3 {x}^{2} - 6 x} = \arctan \frac{{a}^{2} - 4}{3 {a}^{2} - 6 a}$

#### Explanation:

The function $f \left(x\right) = \arctan \frac{{x}^{2} - 4}{3 {x}^{2} - 6 x} = \arctan \frac{{x}^{2} - 4}{3 x \left(x - 2\right)}$ is known to be continuous everywhere except $x = 0$ and $x = 2$ (this could be proved, but it would be a lot of work).

To "evaluate a limit by using continuity" means that you can evaluate the limit of a known continuous function at a point of continuity just by substituting the point into the function. That is, if $f$ is defined near $x = a$ and continuous at $x = a$, then ${\lim}_{x \to a} f \left(x\right) = f \left(a\right)$. Applying this fact to the function $f$ gives:

${\lim}_{x \to a} f \left(x\right) = {\lim}_{x \to a} \arctan \frac{{x}^{2} - 4}{3 {x}^{2} - 6 x} = f \left(a\right) = \arctan \frac{{a}^{2} - 4}{3 {a}^{2} - 6 a}$

when $a \ne 0$ and $a \ne 2$.

This is not part of the given question, but it turns out that, even though $f \left(2\right)$ does not exist and $f$ is not continuous at $x = 2$, the limit ${\lim}_{x \to 2} f \left(x\right)$ does exist in this example. In fact, from L'Hopital's Rule , it can be shown that ${\lim}_{x \to 2} \arctan \frac{{x}^{2} - 4}{3 {x}^{2} - 6 x} = \frac{2}{3}$. Try seeing if you can prove this!