Limit without using L'Hopital's rule.

[cambo86]

Homework Statement

<br /> \lim_{x\rightarrow 1} {\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}}<br />

I can do this very easily using L'Hopital's rule but in the textbook 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?

 

[Curious3141]

Homework Statement

<br /> \lim_{x\rightarrow 1} {\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}}<br />

I can do this very easily using L'Hopital's rule but in the textbook 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.

 

[SammyS]

Homework Statement

<br /> \lim_{x\rightarrow 1} {\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}}<br />I can do this very easily using L'Hopital's rule but in the textbook 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.