Rewriting expression with a limit

**Lepzed** · July 22nd 2012, 11:53 AM

I'm working through a chapter on differentiation and I don't understand how they've rewrote a certain expression:

$\lim_{x \to 0}\frac{\sqrt[3]{x} - \sqrt[3]{0}}{x - 0} = \lim_{x \to 0}\frac{1}{\sqrt[3]{x^2}}$ . I need some step by step help on how they rewrote the left hand part to the right hand part...

**Plato** · July 22nd 2012, 12:01 PM

Originally Posted by Lepzed

I'm working through a chapter on differentiation and I don't understand how they've rewrote a certain expression:

$\lim_{x \to 0}\frac{\sqrt[3]{x} - \sqrt[3]{0}}{x - 0} = \lim_{x \to 0}\frac{1}{\sqrt[3]{x^2}}$ . I need some step by step help on how they rewrote the left hand part to the right hand part...

Just note that $\frac{\sqrt[3]{x} - \sqrt[3]{0}}{x - 0}=\frac{\sqrt[3]{x} }{x }=\frac{\sqrt[3]{x} }{x }\frac{\sqrt[3]{x^2}}{\sqrt[3]{x^2}}=~?$

**Lepzed** · July 22nd 2012, 12:07 PM

Noted and this particular number is chosen to remove the root from the numerator?

**Prove It** · July 22nd 2012, 06:37 PM

Originally Posted by Lepzed

Noted and this particular number is chosen to remove the root from the numerator?

Yes, so that it can easily be seen that you have a denominator approaching 0. What does that tell you about the function?

**Lepzed** · July 22nd 2012, 10:51 PM

That the limit goes to infinity and that function is not differentiable in that point

**Prove It** · July 23rd 2012, 02:16 AM

Originally Posted by Lepzed

That the limit goes to infinity and that function is not differentiable in that point

Correct

**asad** · July 24th 2012, 10:20 PM

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