Question 22 of 70intermediate🔧 ApplyShort Answer2 marks

Find lim⁡x→1f(x)\lim_{x \to 1} f(x), where f(x)={x2−1,x≤1−x2−1,x>1f(x) = \begin{cases} x^2 - 1, & x \leq 1 \\ -x^2 - 1, & x > 1 \end{cases}

Correct Answer

The limit does not exist.

Exercise: EXERCISE 12.1 | Q: 24 | (Page Page 22)
For More Understanding

Explanation

This question involves finding the limit of a piecewise function at x = 1. Following the approach shown in Illustration 9 of the context, we need to evaluate the left-hand limit and right-hand limit separately. For x ≤\leq 1, the function is x² - 1, and for x > 1, the function is -x² - 1. Since the left-hand limit (0) does not equal the right-hand limit (-2), the limit does not exist.

Solution Steps

  1. Left-Hand Limit: For x approaching 1 from the left (x ≤\leq 1), we use f(x) = x² - 1. Therefore, lim(x→1⁻) f(x) = lim(x→1⁻) (x² - 1) = 1² - 1 = 0.

  2. Right-Hand Limit: For x approaching 1 from the right (x > 1), we use f(x) = -x² - 1. Therefore, lim(x→1⁺) f(x) = lim(x→1⁺) (-x² - 1) = -(1)² - 1 = -2.

  3. Conclusion: Since lim(x→1⁻) f(x) = 0 and lim(x→1⁺) f(x) = -2, the left-hand limit is not equal to the right-hand limit. Therefore, lim(x→1) f(x) does not exist.

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