Question

सीमाओं के मान प्राप्त कीजिए : $\lim_{z\rightarrow1}\dfrac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}$

Answer 1

Question:-

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Evaluate $\displaystyle\sf {\lim_{z\to1}\dfrac {z^{\frac {1}{3}}-1}{z^{\frac{1}{6}}-1}.}$

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Answer:-

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$\displaystyle\large\boxed {\sf {\lim_{z\to1}\dfrac {z^{\frac {1}{3}}-1}{z^{\frac{1}{6}}-1}=2}}$

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Solution:-

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We know that,

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$\displaystyle\sf {\lim_{x\to a}\dfrac {x^n-a^n}{x-a}=na^{n-1}\quad\quad\dots (1)}$

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Thus,

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

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

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By (1),

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

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However, $\displaystyle\sf {1^n=1,}$ right?!

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

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

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The shortcut...

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Suppose we are given to evaluate $\displaystyle\sf {\lim_{x\to a}\dfrac {x^p-a^p}{x^q-a^q}.}$

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So,

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$\displaystyle\longrightarrow\sf {\lim_{x\to a}\dfrac {x^p-a^p}{x^q-a^q}=\lim_{x\to a}\dfrac {\left (\dfrac {x^p-a^p}{x-a}\right)}{\left (\dfrac {x^q-a^q}{x-a}\right)}}$

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$\displaystyle\longrightarrow\sf {\lim_{x\to a}\dfrac {x^p-a^p}{x^q-a^q}=\dfrac {\displaystyle\sf{\left (\lim_{x\to a}\dfrac {x^p-a^p}{x-a}\right)}}{\displaystyle\sf{\left (\lim_{x\to a}\dfrac {x^q-a^q}{x-a}\right)}}}$

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$\displaystyle\longrightarrow\sf {\lim_{x\to a}\dfrac {x^p-a^p}{x^q-a^q}=\dfrac {pa^{p-1}}{qa^{q-1}}}$

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$\displaystyle\longrightarrow\sf {\lim_{x\to a}\dfrac {x^p-a^p}{x^q-a^q}=\dfrac {p}{q}\ a^{(p-1)-(q-1)}}$

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$\displaystyle\longrightarrow\sf {\underline {\underline {\lim_{x\to a}\dfrac {x^p-a^p}{x^q-a^q}=\dfrac {p}{q}\ a^{p-q}}}}$

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Special Case - If $\displaystyle\sf {a=1,}$

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$\displaystyle\longrightarrow\sf {\lim_{x\to a}\dfrac {x^p-a^p}{x^q-a^q}=\dfrac {p}{q}\ 1^{p-q}}$

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Remember that $\displaystyle\sf {1^n=1\ !}$

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$\displaystyle\longrightarrow\sf {\underline {\underline {\lim_{x\to a}\dfrac {x^p-a^p}{x^q-a^q}=\dfrac {p}{q}}}}$

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So this formula can directly be applied to such questions.

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And one thing also. Consider that (1).

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$\displaystyle\sf {\lim_{x\to a}\dfrac {x^n-a^n}{x-a}=na^{n-1}}$

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What about if $\displaystyle\sf {a=1\ ?}$

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$\displaystyle\sf {\lim_{x\to 1}\dfrac {x^n-1^n}{x-1}=n(1)^{n-1}}$

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Well, it is false that $\displaystyle\sf {1^n\neq1\ !}$

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$\displaystyle\sf {\underline {\underline {\lim_{x\to 1}\dfrac {x^n-1}{x-1}=n}}}$

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This formula should also be remembered.

Answer 2

2       $\lim_{z\rightarrow1}\dfrac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}$  = 2

Step-by-step explanation:

sajeevanoonampb4du0 से   दूसरा तरीका  

$\lim_{z\rightarrow1}\dfrac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}$

z = 1  प्रयोग करने पर  

= (1- 1)/(1 - 1)

= 0/0

परिभाषित नहीं

Z^(1/3) - 1  

= (Z^(1/6))² - 1²

a² - b² = (a +b)(a - b)

= (Z^(1/6) + 1) (Z^(1/6) - 1)

(Z^(1/3) - 1  )/(Z^(1/6) - 1)

=  (Z^(1/6) + 1) (Z^(1/6) - 1)/(Z^(1/6) - 1)

=   (Z^(1/6) + 1)

z = 1 प्रयोग करने पर  

= 1^(1/6) + 1

= 1 + 1

= 2

$\lim_{z\rightarrow1}\dfrac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}$     = 2

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