Question

lim z^{1/3} - 1 / z^1/6 - 1 z→1

Answer 1

We're asked to evaluate,

$\displaystyle\longrightarrow\sf{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\right)=\,?}$

If we directly take $\sf{z=1}$ we get indeterminate form.

$\displaystyle\longrightarrow\sf{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\right)=\dfrac{1^{\frac{1}{3}}-1}{1^{\frac{1}{6}}-1}}$

$\displaystyle\longrightarrow\sf{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\right)=\dfrac{1-1}{1-1}}$

$\displaystyle\longrightarrow\sf{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\right)=\dfrac{0}{0}}$

Let us divide both the numerator and denominator of the fraction by $\sf{z-1}.$

$\displaystyle\longrightarrow\sf{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\right)=\lim_{z\to1}\left(\dfrac{\frac{z^{\frac{1}{3}}-1}{z-1}}{\frac{z^{\frac{1}{6}}-1}{z-1}}\right)}$

The limit can be distributed to the numerator and the denominator each as,

$\displaystyle\longrightarrow\sf{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\right)=\dfrac{\displaystyle\lim_{z\to1}\left(\frac{z^{\frac{1}{3}}-1}{z-1}\right)}{\displaystyle\lim_{z\to1}\left(\frac{z^{\frac{1}{6}}-1}{z-1}\right)}}$

Or,

$\displaystyle\longrightarrow\sf{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\right)=\dfrac{\displaystyle\lim_{z\to1}\left(\frac{z^{\frac{1}{3}}-1^{\frac{1}{3}}}{z-1}\right)}{\displaystyle\lim_{z\to1}\left(\frac{z^{\frac{1}{6}}-1^{\frac{1}{6}}}{z-1}\right)}\quad\quad\dots(1)}$

Let us recall the formula,

$\displaystyle\longrightarrow\sf{\lim_{x\to a}\left(\dfrac{x^n-a^n}{x-a}\right)=na^{n-1}}$

Thus we have,

$\displaystyle\longrightarrow\sf{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{3}}-1^{\frac{1}{3}}}{z-1}\right)=\dfrac{1}{3}\cdot(1)^{\frac{1}{3}-1}}$

$\displaystyle\longrightarrow\sf{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{3}}-1^{\frac{1}{3}}}{z-1}\right)=\dfrac{1}{3}}$

And,

$\displaystyle\longrightarrow\sf{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{6}}-1^{\frac{1}{6}}}{z-1}\right)=\dfrac{1}{6}\cdot(1)^{\frac{1}{6}-1}}$

$\displaystyle\longrightarrow\sf{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{6}}-1^{\frac{1}{6}}}{z-1}\right)=\dfrac{1}{6}}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\right)=\dfrac{\left(\dfrac{1}{3}\right)}{\left(\dfrac{1}{6}\right)}}$

$\displaystyle\sf{\longrightarrow\sf{\underline{\underline{\lim_{z\to1}\left(\dfrac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\right)=2}}}$

Hence 2 is the answer.