Question 2 Let the function ff be defined as follows: f(x)={3x+1,x<1x2,x≥1f(x) = \begin{cases} 3x + 1, & x < 1 \\ x^2, & x \geq 1 \end{cases} (a) Compute limx→1−f(x)\lim_{x \to 1^-} f(x) (b) Compute limx→1+f(x)\lim_{x \to 1^+} f(x) (c) Does limx→1f(x)\lim_{x \to 1} f(x) exist? Justify your answer. Show solutionHide solution+Question 2 - Solution We are given: f(x)={3x+1,x<1x2,x≥1f(x) = \begin{cases} 3x + 1, & x < 1 \\ x^2, & x \geq 1 \end{cases} (a) Left-hand limit: For x<1x < 1, use f(x)=3x+1f(x) = 3x + 1: limx→1−f(x)=limx→1−(3x+1)=3(1)+1=4\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (3x + 1) = 3(1) + 1 = 4 (b) Right-hand limit: For x≥1x \geq 1, use f(x)=x2f(x) = x^2: limx→1+f(x)=limx→1+x2=12=1\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: limx→1−f(x)=4andlimx→1+f(x)=1⇒limx→1f(x) does not exist\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: limx→1−f(x)=4,limx→1+f(x)=1,limx→1f(x) does not exist\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} }
