One-Sided Limits — Question 2

Question 2

Let the function $f$ be defined as follows: $$f(x) = \begin{cases} 3x + 1, & x < 1 \\ x^2, & x \geq 1 \end{cases}$$

(a) Compute $\lim_{x \to 1^-} f(x)$

(b) Compute $\lim_{x \to 1^+} f(x)$

(c) Does $\lim_{x \to 1} f(x)$ exist? Justify your answer.

Question 2 - Solution

We are given: $$f(x) = \begin{cases} 3x + 1, & x < 1 \\ x^2, & x \geq 1 \end{cases}$$

(a) Left-hand limit: For $x < 1$, use $f(x) = 3x + 1$: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x + 1) = 3(1) + 1 = 4$$

(b) Right-hand limit: For $x \geq 1$, use $f(x) = x^2$: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 = 1^2 = 1$$

(c) Two-sided limit: Since the left and right limits are not equal: $$\lim_{x \to 1^-} f(x) = 4 \quad \text{and} \quad \lim_{x \to 1^+} f(x) = 1 \Rightarrow \lim_{x \to 1} f(x) \text{ does not exist}$$

Final Answer: $$\boxed{ \lim_{x \to 1^-} f(x) = 4, \quad \lim_{x \to 1^+} f(x) = 1, \quad \lim_{x \to 1} f(x) \text{ does not exist} }$$