Limit of a function that has a common denominator

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January 10th 2011, 11:27 PM
evankiefl

Limit of a function that has a common denominator

lim as x goes to 1 of

(x^3 - 1)/(x^4 - 1)

Evaluated as shown above, 0/0 is obtained, which leads me to believe there is a common factor between the divisor and the dividend that can be cancelled so the limit can be evaluated. The problem is, I have no idea how to deal with difference of cubes and difference of 4th degrees. Any help would be much appreciated.
January 10th 2011, 11:33 PM
Failure

Quote:

Originally Posted by evankiefl

lim as x goes to 1 of

(x^3 - 1)/(x^4 - 1)

Evaluated as shown above, 0/0 is obtained, which leads me to believe there is a common factor between the divisor and the dividend that can be cancelled so the limit can be evaluated. The problem is, I have no idea how to deal with difference of cubes and difference of 4th degrees. Any help would be much appreciated.

You can factor out $x-1$ and in both the numerator and the denominator, then cancel it:

$\frac{x^3-1}{x^4-1}=\frac{(x-1)\cdot(x^2+x+1)}{(x-1)\cdot (x^3+x^2+x+1)}=\ldots$
January 11th 2011, 12:03 AM
Prove It

Quote:

Originally Posted by evankiefl

lim as x goes to 1 of

(x^3 - 1)/(x^4 - 1)

Evaluated as shown above, 0/0 is obtained, which leads me to believe there is a common factor between the divisor and the dividend that can be cancelled so the limit can be evaluated. The problem is, I have no idea how to deal with difference of cubes and difference of 4th degrees. Any help would be much appreciated.

You could also use L'Hospital's Rule...

$\displaystyle \lim_{x \to 1}\frac{x^3 - 1}{x^4 - 1} = \lim_{x \to 1}\frac{\frac{d}{dx}(x^3 - 1)}{\frac{d}{dx}(x^4 - 1)}$

$\displaystyle = \lim_{x \to 1}\frac{3x^2}{ 4x^3}$

$\displaystyle = \frac{3}{4}$ .
January 11th 2011, 08:33 AM
Krizalid

$\displaystyle\frac{{{x}^{3}}-1}{{{x}^{4}}-1}=\frac{(x-1)\left( {{x}^{2}}+x+1 \right)}{(x+1)(x-1)\left( {{x}^{2}}+1 \right)}.$
January 11th 2011, 10:04 AM
evankiefl

Thanks everyone, and interesting L'Hopital's Rule technique. What exactly is the rule?
January 11th 2011, 10:45 AM
dwsmith

With L'Hopital's Rule, whenever you have an indeterminant form (see below), you can take the derivative of the numerator, denominator, and then run the limit.

Indeterminant forms:

$\displaystyle\frac{0}{0}, \ \ \frac{\infty}{\infty}, \ \ 0*\infty, \ \ 1^{\infty}, \ \ 0^0, \ \ \infty^0, \ \ \infty - \infty$