Question

Use continuity to evaluate the limit as x approaches 2 of the function arctan(x2-4/3x^2-6x).
I'm stuck; (x+2)/3x using substitution, why does every answer say 2+2/3(2)=2/3?

Answer 1

you are calculating the limit as x -> 2 and not the value at x = 2, because the function's value is not defined at x = 2 -- which is because the (x-2)/(x-2) that you can factor out is undefined at x = 2.

in the limit x -> 2, however, the (x-2)/(x-2) term equals 1 and so the limit becomes (x+2)/3x

at x = 2 that has a precise value of arctan (2+2)/(3*2) = acrctan (2/3) and so the limit has the value arctan (2/3)

Answer 2

\[
\begin{align}
\lim_{x\to 2} \arctan\left(\frac{x^2-4}{3x^2-6x}\right)
& =
\lim_{x\to 2} \arctan\left(\frac{(x+2)(x-2)}{3x(x-2)}\right) \quad \text{Start by factoring numerator \& denominator}\\
& =
\lim_{x\to 2} \arctan\left(\frac{(x+2)}{3x}\right) \quad \text{cancel terms, this function is continuous at $x=2$ so you can now substitute and simplify} \\
& =
\arctan{\left(\frac{((2)+2)}{3(2)}\right)}\\
& =
\arctan{\left(\frac{4}{6}\right)}\\
& =
\arctan{\left(\frac{2}{3}\right)}
\end{align}
\]