15 December 2021

Statement

Is the following function continuous at \(x = 4\) ?

\(g(x) = \frac{x + 1}{2x^2 - 1}\)

Solution

\(g(x) = \frac{x + 1}{2x^2 - 1}\)

Let's start by finding the limit of \(g(x)\) when \(x \to 4\) from both sides.


From the right side

\(\lim\limits_{x \to 4^{+}} g(x)\)

\(= \lim\limits_{x \to 4^{+}} \frac{x + 1}{2x^2 - 1}\)

\(= \lim\limits_{x \to 4^{+}} \frac{5^{+}}{32^{+} - 1}\)

\(= \frac{5}{31}\)


From the left side

\(\lim\limits_{x \to 4^{-}} g(x)\)

\(= \lim\limits_{x \to 4^{-}} \frac{x + 1}{2x^2 - 1}\)

\(= \lim\limits_{x \to 4^{-}} \frac{5^{-}}{32^{-} - 1}\)

\(= \frac{5}{31}\)


Since \(\lim\limits_{x \to 4^{-}} g(x) = \lim\limits_{x \to 4^{+}} g(x)\) and \(g(4) = \frac{5}{31}\) then the function is continous at \(x = 4\).