Question 23 of 70intermediate🔧 ApplyShort Answer2 marks

Evaluate lim⁡x→0f(x)\lim_{x \to 0} f(x), where f(x)={∣x∣x,x≠00,x=0f(x) = \begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}

Correct Answer

To evaluate lim⁡x→0f(x)\lim_{x \to 0} f(x), we find the left-hand limit and right-hand limit separately.

For x<0x < 0: We have ∣x∣=−x|x| = -x, so f(x)=∣x∣x=−xx=−1f(x) = \frac{|x|}{x} = \frac{-x}{x} = -1. Therefore, lim⁡x→0−f(x)=−1\lim_{x \to 0^-} f(x) = -1.

For x>0x > 0: We have ∣x∣=x|x| = x, so f(x)=∣x∣x=xx=1f(x) = \frac{|x|}{x} = \frac{x}{x} = 1. Therefore, lim⁡x→0+f(x)=1\lim_{x \to 0^+} f(x) = 1.

Since the left-hand limit (−1-1) and right-hand limit (11) are different, the limit of f(x)f(x) as xx tends to zero does not exist.

Exercise: EXERCISE 12.1 | Q: 25 | (Page Page 22)
For More Understanding

Explanation

This question follows the same methodology as Illustration 9 in the textbook. The function is piecewise defined, and we need to evaluate the left-hand limit and right-hand limit separately. For negative values of x, |x| = -x, giving f(x) = -1. For positive values of x, |x| = x, giving f(x) = 1. As stated in the Summary section of the context: 'If the right and left hand limits coincide, we call that common value as the limit...Since the left and right hand limits at 0 do not coincide, we say that the limit of the function at 0 does not exist.'

Solution Steps

  1. Step 1: Find left-hand limit: For x < 0, |x| = -x, so f(x) = -x/x = -1. Thus lim(x→0⁻) f(x) = -1

  2. Step 2: Find right-hand limit: For x > 0, |x| = x, so f(x) = x/x = 1. Thus lim(x→0⁺) f(x) = 1

  3. Step 3: Compare limits: Since -1 ≠\neq 1, the limit does not exist

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