Limit without using L'Hopital's rule.

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Homework Help Overview

The discussion centers around evaluating the limit \(\lim_{x\rightarrow 1} {\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}}\) without using L'Hopital's rule, as the problem is presented prior to the introduction of that method in the textbook.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various algebraic manipulations, including the use of the generalized binomial theorem and factoring techniques related to differences of squares and cubes. Questions about the applicability of these methods are raised.

Discussion Status

Several participants have contributed suggestions for alternative approaches to the limit, including algebraic simplifications and factorizations. The discussion is ongoing, with helpful insights being shared, but no consensus has been reached on a single method.

Contextual Notes

The problem is framed within the constraints of not using L'Hopital's rule, prompting participants to consider foundational algebraic techniques instead.

cambo86
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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?
 
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cambo86 said:

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.
 
cambo86 said:

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.
 
Thanks, both responses were very helpful.
 

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