Let's solve the problem step-by-step. The function is defined as: \[ f(x)= \begin{cases} x+1 & \text{if } x < 2 \\ x^2-2 & \text{if } 2 \leq x < 4 \\ \sqrt{x+5} & \text{if } 2 < x < 4 \\ 9 & \text{if } x \geq 4 \end{cases} \] But notice there seems to be a typo: **two pieces are assigned for \(2 < x < 4\)**. Let's assume the correct version is: \[ f(x)= \begin{cases} x+1 & x < 2 \\ x^2-2 & x = 2 \\ \sqrt{x+5} & 2 < x < 4 \\ 9 & x \geq 4 \end{cases} \] Let's solve the given limits: --- ### (a) \(\lim_{x \to 2^-} f(x)\) - For \(x < 2\), \(f(x) = x + 1\) - As \(x \to 2^-\), \(f(x) \to 2 + 1 = 3\) **Answer:** \[ \boxed{3} \] --- ### (b) \(\lim_{x \to 2^+} f(x)\) - For \(2 < x < 4\), \(f(x) = \sqrt{x+5}\) - As \(x \to 2^+\), \(f(x) \to \sqrt{2+5} = \sqrt{7}\) **Answer:** \[ \boxed{\sqrt{7}} \] --- ### (c) \(\lim_{x \to 2} f(x)\) - Left limit = 3, right limit = \(\sqrt{7}\) - Limits are **not equal**, so the limit **does not exist**. **Answer:** \[ \boxed{\text{Does not exist}} \] --- ### (d) \(\lim_{x \to 4^-} f(x)\) - For \(2 < x < 4\), \(f(x) = \sqrt{x+5}\) - As \(x \to 4^-\), \(f(x) \to \sqrt{4+5} = \sqrt{9} = 3\) **Answer:** \[ \boxed{3} \] --- ### (e) \(\lim_{x \to 4^+} f(x)\) - For \(x \geq 4\), \(f(x) = 9\) - As \(x \to 4^+\), \(f(x) = 9\) **Answer:** \[ \boxed{9} \] --- ### (f) \(\lim_{x \to 4} f(x)\) - Left-hand limit = 3, right-hand limit = 9 - Limits are **not equal**, so the limit **does not exist**. **Answer:** \[ \boxed{\text{Does not exist}} \] --- ### Final Answers (a) \(3\) (b) \(\sqrt{7}\) (c) Does not exist (d) \(3\) (e) \(9\) (f) Does not exist Let me know if you need any further clarifications!

Certainly! Let's carefully analyze and compute each part step-by-step, including detailed calculations and reasoning. --- ### Given piecewise function: \[ f(x) = \begin{cases} x + 1 & x < 2 \\ x^2 - 2 & x = 2 \\ \sqrt{x + 5} & 2 < x < 4 \\ 9 & x \geq 4 \end{cases} \] (Note: The function appears to have some inconsistency in the original image, but based on typical piecewise functions and the points involved, this is a logical interpretation.) --- ## (a) \(\displaystyle \lim_{x \to 2^-} f(x)\) (Limit from the left of 2) **Step 1:** Identify the relevant piece for \(x < 2\): \[ f(x) = x + 1 \] **Step 2:** Compute the limit as \(x \to 2^-\): \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x + 1) \] Since \(x + 1\) is continuous, just substitute \(x = 2\): \[ = 2 + 1 = 3 \] **Final answer:** \[ \boxed{3} \] --- ## (b) \(\displaystyle \lim_{x \to 2^+} f(x)\) (Limit from the right of 2) **Step 1:** Identify the relevant piece for \(x > 2\): \[ f(x) = \sqrt{x + 5} \] **Step 2:** Compute the limit as \(x \to 2^+\): \[ \lim_{x \to 2^+} \sqrt{x + 5} \] Substitute \(x = 2\): \[ \sqrt{2 + 5} = \sqrt{7} \] **Final answer:** \[ \boxed{\sqrt{7}} \] --- ## (c) \(\displaystyle \lim_{x \to 2} f(x)\) **Step 1:** Recall the left and right limits: - Left limit: 3 - Right limit: \(\sqrt{7}\) Since these are not equal: \[ \boxed{\text{Limit does not exist}} \] --- ## (d) \(\displaystyle \lim_{x \to 4^-} f(x)\) (Limit approaching 4 from left) **Step 1:** For \(2 < x < 4\), \(f(x) = \sqrt{x + 5}\) **Step 2:** Compute the limit as \(x \to 4^-\): \[ \lim_{x \to 4^-} \sqrt{x + 5} \] Substitute \(x = 4\): \[ \sqrt{4 + 5} = \sqrt{9} = 3 \] **Final answer:** \[ \boxed{3} \] --- ## (e) \(\displaystyle \lim_{x \to 4^+} f(x)\) (Limit approaching 4 from right) **Step 1:** For \(x \geq 4\), \(f(x) = 9\) **Step 2:** Since the function is constant (or at least approaches a constant) as \(x \to 4^+\): \[ \lim_{x \to 4^+} f(x) = 9 \] **Final answer:** \[ \boxed{9} \] --- ## (f) \(\displaystyle \lim_{x \to 4} f(x)\) **Step 1:** Compare the left and right limits: - Left limit: 3 - Right limit: 9 Since they are not equal: \[ \boxed{\text{Limit does not exist}} \] --- ## **Summary of all parts:** | Part | Limit/Value | Explanation | |---------|-------------------|------------------------------| | (a) | \(3\) | From \(x<2\), \(f(x)=x+1\), limit as \(x \to 2^-\) is 3 | | (b) | \(\sqrt{7}\) | From \(x>2\), \(f(x)=\sqrt{x+5}\), limit as \(x \to 2^+\) is \(\sqrt{7}\) | | (c) | Does not exist | Left limit 3, right limit \(\sqrt{7}\) are not equal | | (d) | \(3\) | From \(x<4\), \(f(x)=\sqrt{x+5}\), limit as \(x \to 4^-\) is 3 | | (e) | \(9\) | For \(x \geq 4\), \(f(x)=9\), limit as \(x \to 4^+\) is 9 | | (f) | Does not exist | Left limit 3, right limit 9 are not equal | --- Please let me know if you'd like further clarification or additional details!