Question

If $f(x) = \begin{cases} x^2 - 1 & \text{if } x \ge 2 \\ x + 1 & \text{if } x<2 \end{cases}$, then $\lim_{x \to 2^+} f(x) + \lim_{x \to 2^-} f(x) = $

• A. 7
• B. 5
• C. 6
• D. 9

Hint:

For piecewise functions, the most critical step is selecting the correct "piece" of the equation based on whether you are evaluating a left-sided limit ($x<c$) or a right-sided limit ($x>c$).

Answer 1

The Correct Option is C

To determine the value of \(\lim_{x \to 2^+} f(x) + \lim_{x \to 2^-} f(x)\), we need to evaluate the left-hand limit and the right-hand limit of the function \( f(x) \) as \( x \) approaches 2.

1. Right-Hand Limit (\( x \to 2^+ \)):

Since this is the right-hand limit, \( x \) approaches 2 from values greater than 2. Therefore, use the definition of \( f(x) \) for \( x \ge 2 \):

\(f(x) = x^2 - 1\)

Substitute \( x = 2 \):

\(f(2) = 2^2 - 1 = 3\)

Thus, \(\lim_{x \to 2^+} f(x) = 3\).

1. Left-Hand Limit (\( x \to 2^- \)):

Since this is the left-hand limit, \( x \) approaches 2 from values less than 2. Therefore, use the definition of \( f(x) \) for \( x < 2 \):

\(f(x) = x + 1\)

Substitute \( x = 2 \):

\(f(2) = 2 + 1 = 3\)

Thus, \(\lim_{x \to 2^-} f(x) = 3\).

Now, add the two limits:

\(\lim_{x \to 2^+} f(x) + \lim_{x \to 2^-} f(x) = 3 + 3 = 6\)

Therefore, the correct answer is 6.