Grade 12 calculas limits question

**koolaid123** · February 13th 2013, 05:39 PM

lim x->1

solve algebraeically...ive tried it a bunch of times and always seem to head nowhere

**Plato** · February 13th 2013, 07:00 PM

Originally Posted by koolaid123

lim x->1

solve algebraeically...ive tried it a bunch of times and always seem to head nowhere

Simple algebra will show that:
$\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}$ .

**Soroban** · February 13th 2013, 07:10 PM

Hello, koolaid123!

$\lim_{x\to1}\frac{\sqrt[3]{x} - 1}{\sqrt{x}-1}$

$\text{Multiply by }\frac{\sqrt{x}+1}{\sqrt{x}+1}\,\text{ and }\,\frac{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}$

. . $\lim_{x\to1}\frac{\sqrt[3]{x} - 1}{\sqrt{x}-1}\cdot \frac{\sqrt{x}+1}{\sqrt{x}+1} \cdot \frac{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}$

. . $=\;\lim_{x\to1} \frac{(\sqrt[3]{x} - 1)(\sqrt[3]{x^2} + \sqrt[3]{x} + 1)}{(\sqrt{x}-1)(\sqrt{x}+1)} \cdot\frac{\sqrt{x}+1}{\sqrt[3]{x^2} + \sqrt[3]{x}+1}$

. . $=\;\lim_{x\to1}\frac{x-1}{x-1}\cdot\frac{\sqrt{x}+1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}$

. . $=\;\lim_{x\to1}\frac{\sqrt{x}+1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1} \;=\;\frac{1+1}{1+1+1} \;=\;\frac{2}{3}$

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