Question 10 of 70intermediate🔧 ApplyNumerical1 mark

Evaluate the following limits in Exercises 1 to 22. lim⁡z→1z13−1z16−1\lim_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}

Correct Answer

2

Exercise: EXERCISE 12.1 | Q: 10 | (Page Page 21)
For More Understanding

Explanation

The limit is evaluated by substituting x = z1/6z^{1/6}, which transforms the expression into a standard limit form. Using the theorem lim_{x→a} (xnx^{n} - ana^{n})/(x - a) = n an−1a^{n-1} (provided in the context), we obtain the result 2.

Solution Steps

  1. Step 1: Let x = z1/6z^{1/6}. Then z1/3z^{1/3} = (z1/6z^{1/6})^2 = x2x^{2}, and as z→1, x→1.

  2. Step 2: The limit becomes lim_{x→1} (x2x^{2} - 1)/(x - 1).

  3. Step 3: Apply the standard limit: lim_{x→a} (xnx^{n} - ana^{n})/(x - a) = n an−1a^{n-1}. Here n=2, a=1, so limit = 2·111^{1} = 2.

  4. Step 4: Alternatively, factor (x2x^{2}-1) = (x-1)(x+1), cancel (x-1), and evaluate lim_{x→1} (x+1) = 2.

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