Question 21 of 70intermediate🔧 ApplyShort Answer2 marks

Find lim⁡x→0f(x)\lim_{x \to 0} f(x) and lim⁡x→1f(x)\lim_{x \to 1} f(x), where f(x)={2x+3,x≤03(x+1),x>0f(x) = \begin{cases} 2x + 3, & x \leq 0 \\ 3(x + 1), & x > 0 \end{cases}

Correct Answer

(i) lim⁡x→0f(x)=3\lim_{x \to 0} f(x) = 3

(ii) lim⁡x→1f(x)=6\lim_{x \to 1} f(x) = 6

Exercise: EXERCISE 12.1 | Q: 23 | (Page Page 22)
For More Understanding

Explanation

This question tests the concept of limits of piecewise functions. The student needs to evaluate left-hand and right-hand limits at the point where the function definition changes (x = 0), and also at a point where only one definition applies (x = 1). The textbook demonstrates this approach in Illustration 9, where limits are evaluated by checking values from both sides of the point.

Solution Steps

  1. Step 1: For lim⁡x→0f(x)\lim_{x \to 0} f(x), find Left-hand limit: When x→0−x \to 0^-, we use f(x)=2x+3f(x) = 2x + 3 (since x≤0x \leq 0). So lim⁡x→0−f(x)=2(0)+3=3\lim_{x \to 0^-} f(x) = 2(0) + 3 = 3.

  2. Step 2: Find Right-hand limit: When x→0+x \to 0^+, we use f(x)=3(x+1)f(x) = 3(x + 1) (since x>0x > 0). So lim⁡x→0+f(x)=3(0+1)=3\lim_{x \to 0^+} f(x) = 3(0 + 1) = 3.

  3. Step 3: Since lim⁡x→0−f(x)=lim⁡x→0+f(x)=3\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = 3, therefore lim⁡x→0f(x)=3\lim_{x \to 0} f(x) = 3.

  4. Step 4: For lim⁡x→1f(x)\lim_{x \to 1} f(x), note that x=1>0x = 1 > 0, so we use f(x)=3(x+1)f(x) = 3(x + 1) for both sides.

  5. Step 5: lim⁡x→1f(x)=lim⁡x→13(x+1)=3(1+1)=6\lim_{x \to 1} f(x) = \lim_{x \to 1} 3(x + 1) = 3(1 + 1) = 6.

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