calculating limit: limit_{x -> 1} (x - 1)/(x^n - 1) [Can this be done without l'hopital's?]

[Qwertyuiop[]]

[imath]\lim _{x\to 1}\left(\frac{x-1}{x^n-1}\right)[/imath] , i used l'hopital's rule for this, i got [imath]\frac{1}{n}[/imath]. Is there another way of doing it that doesn't involve l'hopital's rule? Also what are the conditions for using the l'hopital's rule? I know it can be used whenever we have an indeterminate form of [imath]\frac{0}{0}[/imath] or [imath]_{\frac{\infty }{_{\infty \:}}\:}[/imath]. Are there any other conditions?

 

[topsquark]

[imath]\lim _{x\to 1}\left(\frac{x-1}{x^n-1}\right)[/imath] , i used l'hopital's rule for this, i got [imath]\frac{1}{n}[/imath]. Is there another way of doing it that doesn't involve l'hopital's rule? Also what are the conditions for using the l'hopital's rule? I know it can be used whenever we have an indeterminate form of [imath]\frac{0}{0}[/imath] or [imath]_{\frac{\infty }{_{\infty \:}}\:}[/imath]. Are there any other conditions?

Well, the functions in the numerator and denominator have to be smooth near the limit point.

-Dan

 

[lev888]

[imath]\lim _{x\to 1}\left(\frac{x-1}{x^n-1}\right)[/imath] , i used l'hopital's rule for this, i got [imath]\frac{1}{n}[/imath]. Is there another way of doing it that doesn't involve l'hopital's rule? Also what are the conditions for using the l'hopital's rule? I know it can be used whenever we have an indeterminate form of [imath]\frac{0}{0}[/imath] or [imath]_{\frac{\infty }{_{\infty \:}}\:}[/imath]. Are there any other conditions?

Factor the denominator.

 

[pka]

[imath]\lim _{x\to 1}\left(\frac{x-1}{x^n-1}\right)[/imath] , i used l'hopital's rule for this, i got [imath]\large\bf\frac{1}{n}~\Large\color{blue}\checkmark[/imath]. Is there another way of doing it that doesn't involve l'hopital's rule? Also what are the conditions for using the l'hopital's rule? I know it can be used whenever we have an indeterminate form of [imath]\frac{0}{0}[/imath] or [imath]_{\frac{\infty }{_{\infty \:}}\:}[/imath]. Are there any other conditions?

Why do you think that [imath]\bf\dfrac{1}{n}[/imath] is not the correct limit? It is the limit!

[imath][/imath][imath][/imath]

 

[Steven G]

Well, the functions in the numerator and denominator have to be smooth near the limit point.

-Dan

Dan,
Please define smooth for the Calculus student.
Steven

 

[Steven G]

I agree with pka. Your method is fine!

 

[Dr.Peterson]

Why do you think that [imath]\bf\dfrac{1}{n}[/imath] is not the correct limit? It is the limit!

[imath][/imath][imath][/imath]

That wasn't the issue here; he knows it's the limit, but was asking about other methods:

[imath]\lim _{x\to 1}\left(\frac{x-1}{x^n-1}\right)[/imath] , i used l'hopital's rule for this, i got [imath]\frac{1}{n}[/imath]. Is there another way of doing it that doesn't involve l'hopital's rule?

I second answer #3: factor the denominator. Or, more or less equivalently, use long division, or other concepts, to express the reciprocal, [imath]\dfrac{x^n-1}{x-1}[/imath], as a summation (series).

A colleague the other day told about a student early in a calculus class, who had taken calculus before and just used L'Hopital for every limit. The question was, why stop him? Once you have the powerful tool, you don't really need to know the special-purpose methods. But I do think it can be a good way to stretch your mind.

On the other hand, such limits are used to define the derivative, so using L'Hopital for them is almost circular reasoning (and is, if you are doing a limit from which the derivative formula you are using is derived).