# Limit without using L'Hopital's rule.

cambo86

## Homework Statement

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

I can do this very easily using L'Hopital's rule but in the text book I'm going through it is a problem given before L'Hopital's rule is taught. Is there a way of doing this without using L'Hopita'ls rule?

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Homework Helper

## Homework Statement

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

I can do this very easily using L'Hopital's rule but in the text book I'm going through it is a problem given before L'Hopital's rule is taught. Is there a way of doing this without using L'Hopita'ls rule?

One way is to let ##x = (y+1)^2##, after which the numerator can be simplified with the generalised binomial theorem.

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## Homework Statement

$$\lim_{x\rightarrow 1} {\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}}$$I can do this very easily using L'Hopital's rule but in the text book I'm going through it is a problem given before L'Hopital's rule is taught. Is there a way of doing this without using L'Hopita'ls rule?
Do you know how to factor the difference of squares and the difference cubes ?

$(a-b)(a+b)=a^2-b^2$

$(c-d)(c^2+cd+d^2)=c^3-d^3$

Therefore, $\ \ (\sqrt{x\,}-1)(\sqrt{x\,}+1)=x-1$

and $\ \ (\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)=x-1\ .$

These can be used to cancel x-1 in the numerator & denominator.

cambo86
Thanks, both responses were very helpful.