# Math Help - Does the limit exist?

1. ## Does the limit exist?

limit as x approaches 1 of: (x^4-1)/(x^3-1)

I factored the top to be (x^2-1)(x^2+1) and the bottom using difference of cubes which I got to be (x-1)(x^2+x+1)

I am stuck at this point because nothing seems to cancel out and I dont see any more factors. So does it not exist?

2. ## Re: Does the limit exist?

Originally Posted by jmanna98
\displaystyle \begin{align*} \frac{x^4 - 1}{x^3 - 1} &= \frac{(x^2 - 1)(x^2 + 1)}{(x - 1)(x^2 + x + 1)} \\ &= \frac{(x - 1)(x + 1)(x^2 + 1)}{(x - 1)(x^2 + x + 1)} \\ &= \frac{(x + 1)(x^2 + 1)}{x^2 + x + 1} \end{align*}
Now make $\displaystyle x \to 1$.