Question
The function f is defined by \( f(x)= \frac{x^{2}-4}{x-2} \) for \(x \neq 2\).
(a) Find \(lim_{x\to 2}\)f(x).
(b) Is f continuous at x=2? Justify your answer.
(c) The function g is defined by
\(g(x) =\begin{cases}
f(x)& \text{ if } x\neq 2 \\
k& \text{ if } x= 2
\end{cases}\).
Find the value of k such that g is continuous at x=2.
▶️Answer/Explanation
(a) We can factor the numerator and simplify the expression:
\(\displaystyle \lim_{x \to 2}f(x) = \displaystyle \lim_{x \to 2}\frac{x^{2}-4}{x-2} =\displaystyle \lim_{x \to 2}\frac{(x+2)(x-2)}{(x-2)}=\displaystyle \lim_{x \to 2}(x+2) = 4\)
(b) No, f is not continuous at x=2 because f(2) is undefined.
(c) For g to be continuous at x=2, we need \(lim_{x\to 2}\)g(x) = g(2).
From part (a), we know that \(lim_{x\to 2}\)g(x) =\(lim_{x\to 2}\)f(x) = 4.
Therefore, we must have g(2)=k=4.