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Exercise 12.1 · Q10

Q.lim⁡z→1z1/3−1z1/6−1\lim_{z\to 1}\dfrac{z^{1/3} - 1}{z^{1/6} - 1}

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Both top and bottom vanish at z=1z=1, giving the 0/00/0 form. The substitution t=z1/6t=z^{1/6} turns it into a difference of squares that cancels, and the limit equals 22.

Direct substitution fails: at z=1z=1 the numerator z1/3−1z^{1/3}-1 and the denominator z1/6−1z^{1/6}-1 are both 00, so we have the indeterminate form 00\dfrac{0}{0}. We remove the common vanishing factor by a substitution.

Step 1 — Substitute to clear the fractional powers.

Let t=z1/6t=z^{1/6}. As z→1z\to 1, t→1t\to 1. Also z1/3=(z1/6)2=t2z^{1/3}=(z^{1/6})^2=t^2, so

lim⁡z→1z1/3−1z1/6−1=lim⁡t→1t2−1t−1.\lim_{z\to 1}\frac{z^{1/3}-1}{z^{1/6}-1}=\lim_{t\to 1}\frac{t^2-1}{t-1}.

Step 2 — Factor the numerator.

t2−1t^2-1 is a difference of squares: t2−1=(t−1)(t+1)t^2-1=(t-1)(t+1). Hence

t2−1t−1=(t−1)(t+1)t−1=t+1(t≠1).\frac{t^2-1}{t-1}=\frac{(t-1)(t+1)}{t-1}=t+1\quad(t\neq 1).

Step 3 — Evaluate. …

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