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

[Guest]

[tex]\lim_{x \to 1} \frac{(x^4)-1}{(x^3)-1}[/tex]

[Math Tutor]

[tex]\lim_{x \to 1} \frac{(x^4)-1}{(x^3)-1} = \lim_{x \to 1} \frac{(x^2-1)(x^2+1)}{(x-1)(x^2+x+1)}=\\=\lim_{x \to 1} \frac{(x-1)(x+1)(x^2+1)}{(x-1)(x^2+x+1)} = \lim_{x \to 1} \frac{(x+1)(x^2+1)}{(x^2+x+1)} = \frac{(1+1)(1+1)}{1+1+1} = \frac{4}{3}[/tex]

[Guest]

Notice that this is an "undetermined" situation because both numerator and denominator go to 0. Because they become 0 we know immediately that we can factor out an "x- 1". That should have been your first thought.

[Guest]

As was said before, this fraction is "indeterminate" at x= 1 because both numerator and denominator are 0 there. That means that each has x- 1 as a factor. In fact,
[tex]x^4- 1= (x- 1)(x^3+ x^2+ x+ 1)[/tex]
[tex]x^3- 1= (x- 1)(x^2+ x+ 1)[/tex].

So [tex]\frac{x^4- 1}{x^3- 1}= \frac{(x- 1)(x^3+ x^2+ x+ 1)}{(x- 1)(x^2+ x+ 1)}[/tex].

As long as x is not 1, that is equal to
[tex]\frac{x^3+ x^2+ x+ 1}{x^2+ x+ 1}[/tex]
and, for x= 1 that is [tex]\frac{4}{3}[/tex].

[Math Tutor]

Yes, this second solution is better.