Question 4 Let the function f(x)f(x) be defined as: f(x)={x2−4,x<12x+1,x≥1f(x) = \begin{cases} x^2 - 4, & x < 1 \\ 2x + 1, & x \geq 1 \end{cases} (a) Find limx→1−f(x)\lim_{x \to 1^-} f(x) (b) Find limx→1+f(x)\lim_{x \to 1^+} f(x) (c) Does limx→1f(x)\lim_{x \to 1} f(x) exist? Explain. Show solutionHide solution+Question 4 - Solution We are given: f(x)={x2−4,x<12x+1,x≥1f(x) = \begin{cases} x^2 - 4, & x < 1 \\ 2x + 1, & x \geq 1 \end{cases} (a) Left-hand limit: For x<1x < 1, we use f(x)=x2−4f(x) = x^2 - 4: limx→1−f(x)=12−4=−3\lim_{x \to 1^-} f(x) = 1^2 - 4 = -3 limx→1−f(x)=−3\boxed{\lim_{x \to 1^-} f(x) = -3} (b) Right-hand limit: For x≥1x \geq 1, we use f(x)=2x+1f(x) = 2x + 1: limx→1+f(x)=2(1)+1=3\lim_{x \to 1^+} f(x) = 2(1) + 1 = 3 limx→1+f(x)=3\boxed{\lim_{x \to 1^+} f(x) = 3} (c) Two-sided limit: limx→1−f(x)=−3andlimx→1+f(x)=3\lim_{x \to 1^-} f(x) = -3 \quad \text{and} \quad \lim_{x \to 1^+} f(x) = 3 Since the one-sided limits are not equal, the overall limit does not exist: limx→1f(x) does not exist\boxed{\lim_{x \to 1} f(x) \text{ does not exist}}
