Limit without using L'Hopital's rule.

  • Thread starter cambo86
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  • #1
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Homework Statement


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

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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Answers and Replies

  • #2
Curious3141
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Homework Statement


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

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.
 
  • #3
SammyS
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Homework Statement

[tex]
\lim_{x\rightarrow 1} {\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}}
[/tex]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 ?

[itex](a-b)(a+b)=a^2-b^2[/itex]

[itex](c-d)(c^2+cd+d^2)=c^3-d^3[/itex]

Therefore, [itex]\ \ (\sqrt{x\,}-1)(\sqrt{x\,}+1)=x-1[/itex]

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

These can be used to cancel x-1 in the numerator & denominator.
 
  • #4
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Thanks, both responses were very helpful.
 

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