Question

If f(x) =\(\begin{cases}\frac{x^2-9}{x-3}, x≠3 \\ 5, x=3 \end {cases}\) then f(x):

• A. is continuous at x = 3
• B. has removable discontinuity at x =3
• C. has irremovable discontinuity at x=3
• D. continuous at every real number

Answer 1

The Correct Option is B

The problem involves determining the continuity of the function \( f(x) = \begin{cases}\frac{x^2-9}{x-3},& x\neq3 \\ 5,& x=3 \end{cases} \). We need to check if there is a removable discontinuity at \( x=3 \).

Step 1: Simplify the expression for \( x \neq 3 \):

\(\frac{x^2-9}{x-3}\) can be simplified by recognizing that \( x^2-9 \) is a difference of squares: \( (x-3)(x+3) \). The expression becomes:

\(\frac{(x-3)(x+3)}{x-3}=x+3\), for \( x \neq 3 \).

Step 2: Investigate the limit as \( x \to 3 \):

\(\lim_{x \to 3}(x+3)=6\).

Step 3: Evaluate the function at \( x=3 \):

Given \( f(3)=5 \).

Conclusion: Since \(\lim_{x \to 3} f(x)=6\) does not equal \( f(3)=5 \), there is a discontinuity at \( x=3 \).

However, since the limit exists, the discontinuity is removable. Thus, the correct description is that the function has a removable discontinuity at \( x=3 \).