Question

limit x➡️1 Z^1/3-1÷Z^1/6-1

I don't need rubbish​

Answer 1

Question :

$\red\bigstar\sf{lim_{x \to 1} \: \dfrac{z^(dfrac{1}{3}) -1}{z^(\dfrac{1}{6}) -1}}$

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To find :

➡$\bf{ the \: value \: of \: lim_{x \to 0}=?}$

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Solution :

$\\ \implies\sf{lim_{x \to 1} \: \dfrac{z^({\dfrac{1}{3}})-1}{z^({\dfrac{1}{6}})-1}}$

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By using L -Hospital rule:

$\\ \implies\sf{lim_{ x \to 1 } \: \dfrac{z^({\dfrac{1}{3})}} {z^({\dfrac{1}{6}})-1}}$

$\\ \implies\sf{lim_{x \to 1} \: \dfrac{z^({\dfrac{1}{3})}}{z^({\dfrac{1}{6}})-1}}$

$\\ \implies\sf{lim_{x \to 1} \: \dfrac{dy}{dx}\dfrac{z^({\dfrac{1}{3}) -1}}{z^({\dfrac{1}{6}})-1}}$

$\\ \implies\sf{lim_{x \to 1} \: \dfrac{(\dfrac{1}{3})z^{(\dfrac{-2}{3}})-0}{(\dfrac{1}{6})z^{(\dfrac{-5}{6})}}}$

$\\ \implies\sf{lim_{ x \to 1 } \: 2 z^{\dfrac{-2}{3}-\dfrac{5}{6}}}$

$\\ \implies\sf{lim_{ x \to 1} \: 2 z^{\dfrac{-2}{3}-\dfrac{+5}{6}}}$

$\\ \implies\sf{lim_{x \to 1 } \: 2z^{\dfrac{-4-5}{6}}}$

$\\ \implies\sf{lim_{ x \to 1 }\: 2z^{\dfrac{-3}{2}}}$

Put x=1 in this term we get

$\\ \implies\sf{lim_{x \to 1} 2 \times 1^{\bigg({\dfrac{-3}{2}\bigg)}}}$

$\\ \implies\sf{lim_{x \to 1} 2}$

The answer will be 2