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Evaluate the following limits \[\] $ \lim\limits_{\large z \to 1} \large\frac{\large z^{\large\frac{1}{3}}-1}{z^{\large\frac{1}{6}}-1}$

$\begin{array}{1 1}2 \\ 3\\ 4\\ 0 \end{array} $

1 Answer

$ \lim\limits_{z \to 1} \large\frac{z^{\Large\frac{1}{3}}-1}{z^{\Large\frac{1}{6}}-1}$
At $ z = 1$ the value of the given fraction takes the form $ \large\frac{0}{0}.$
Put $ z^{\large\frac{1}{6}}=x$ so that $ z \rightarrow 1$ as $ x \rightarrow 1$.
Accordingly, $ \lim\limits_{z \to 1} \large\frac{z^{\Large\frac{1}{3}}-1}{z^{\Large\frac{1}{6}}-1}$ $ = \lim\limits_{z \to 1} \large\frac{x^2-1}{x-1}$
$ = \lim\limits_{z \to 1} \large\frac{x^2-1^2}{x-1}$
$ = 2.1^{2-1} \quad \quad \quad \bigg[ \lim\limits_{x \to a} \large\frac{x^n-a^n}{x-a} = na^{n-1} \bigg]$
$ = 2$
$ \therefore \lim\limits_{z \to 1} \large\frac{z^{\Large\frac{1}{3}}-1}{z^{\Large\frac{1}{6}}-1}$$=2$
answered Apr 3, 2014 by thanvigandhi_1
 

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