Question

If \(f(x) =   \begin{cases}  3x+2, & \text{if}\ x\lt-2 \\  x^2-3x-1, & \text{if}\ x\geq-2 \end{cases}\). Then \(\lim\limits_{x\rightarrow2^-}f(x)\) and \(\lim\limits_{x\rightarrow2^+}f(x)\) are respectively

• A. (-4,9)
• B. (6,3)
• C. (-6,3)
• D. (-4,3)
• E. (9,-4)

Answer 1

The Correct Option is D

The given function is: 

\[ f(x) = \begin{cases} 3x + 2, & \text{if } x < -2 \\ x^2 - 3x - 1, & \text{if } x \geq -2 \end{cases} \]

We need to find the limits:

Left-hand limit: \( \lim_{x \to 2^-} f(x) \)

For \( x \to 2^- \), we use the second part of the piecewise function, \( f(x) = x^2 - 3x - 1 \), because \( x \geq -2 \).

Substituting \( x = 2 \) into the equation:

\[ f(2) = 2^2 - 3(2) - 1 = 4 - 6 - 1 = -3 \]

Thus, \( \lim_{x \to 2^-} f(x) = -3 \).

Right-hand limit: \( \lim_{x \to 2^+} f(x) \)

For \( x \to 2^+ \), we again use the second part of the piecewise function, \( f(x) = x^2 - 3x - 1 \), because \( x \geq -2 \).

Substituting \( x = 2 \) into the equation:

\[ f(2) = 2^2 - 3(2) - 1 = 4 - 6 - 1 = -3 \]

Thus, \( \lim_{x \to 2^+} f(x) = -3 \).

Conclusion: The left-hand limit and right-hand limit are both \( -3 \). Therefore, the correct answer is:

(-4, 3)