Evaluate the following limit-lim z - 1 z 1-3-1- z 1-6-1

Here z → 1limz^1/3 1/z^1/6 1 [ 0/0form] = z → 1lim(z^1/6)^2 (1)^2/z^1/6 1 = z → 1lim(z^1/6+1) (z^1/6 1)/(z^1/6 1) = z → 1lim z^1/6+ 1 = (1)^1/6+1 = 1 + 1 = 2 ...

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Byju's Answer

Standard XI

Mathematics

Fundamental Laws of Logarithms

Evaluate the ...

Question

Evaluate the following limit:
$lim z \to 1 \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}$

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Solution

Here $lim z \to 1 \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1} [\frac{0}{0} form]$

= $lim z \to 1 \frac{(z^{\frac{1}{6}})^{2} - (1)^{2}}{z^{\frac{1}{6}} - 1}$

= $lim z \to 1 \frac{(z^{\frac{1}{6}} + 1) (z^{\frac{1}{6}} - 1)}{(z^{\frac{1}{6}} - 1)}$

= $lim z \to 1 z^{\frac{1}{6}} + 1 = (1)^{\frac{1}{6}} + 1$

= 1 + 1 = 2.

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Similar questions

Q. Evaluate the given limit :

lim z \to 1 \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}

lim z \to 1 \frac{z^{1_{/ 3}} - 1}{z^{1_{/ 6}} - 1}

Q. If

z_{1} = 2 - i, z_{2} = 1 + i, find |\frac{z_{1} + z_{2} + 1}{z_{1} - z_{2} + i}|

Q. If z₁ is a complex number other than −1 such that

|z_{1}| = 1

and

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, then show that the real parts of z₂ is zero.

Q. If z₁, z₂, z₃ are complex numbers such that

|z_{1}| = |z_{2}| = |z_{3}| = |\frac{1}{z_{1}} + \frac{1}{z_{2}} + \frac{1}{z_{3}}| = 1

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Evaluate the following limit: limz→1z13−1z16−1

Here limz→1z13−1z16−1[00form] = limz→1(z16)2−(1)2z16−1 = limz→1(z16+1)(z16−1)(z16−1) = limz→1 z16+1=(1)16+1 = 1 + 1 = 2.

Evaluate the following limit:
$lim z \to 1 \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1}$