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

Step 1: Understanding the Concept:
The problem requires evaluating the right-hand limit (RHL) and the left-hand limit (LHL) of a piecewise-defined function at the point where its definition changes ($x=2$), and then summing the two limit values.

Step 2: Key Formula or Approach:
- To find $\lim_{x \to 2^+} f(x)$ (RHL), use the piece of the function defined for $x>2$. - To find $\lim_{x \to 2^-} f(x)$ (LHL), use the piece of the function defined for $x<2$. - Substitute $x=2$ into these respective pieces to calculate the limits, then add them.

Step 3: Detailed Explanation:
1. Evaluate the Right-Hand Limit (RHL):
We are approaching 2 from values greater than 2 ($x \to 2^+$).
For $x \ge 2$, the function is defined as $f(x) = x^2 - 1$.
Therefore, $\lim_{x \to 2^+} f(x) = \lim_{x \to 2} (x^2 - 1)$.
Substituting $x=2$ directly gives:
$\text{RHL} = (2)^2 - 1 = 4 - 1 = 3$.
2. Evaluate the Left-Hand Limit (LHL):
We are approaching 2 from values less than 2 ($x \to 2^-$).
For $x<2$, the function is defined as $f(x) = x + 1$.
Therefore, $\lim_{x \to 2^-} f(x) = \lim_{x \to 2} (x + 1)$.
Substituting $x=2$ directly gives:
$\text{LHL} = 2 + 1 = 3$.
3. Calculate the sum:
The question asks for the sum of these two limits:
Sum $= \text{RHL} + \text{LHL}$
Sum $= 3 + 3 = 6$.

Step 4: Final Answer:
The sum of the limits is 6.