- #1
shamieh
- 539
- 0
lim x--> 1
\(\displaystyle
\frac{x^4 - 3x^3 + 3x^2 - x}{x^4 - 2x^3 + 2x - 1}\)I got \(\displaystyle \frac{0}{6} = 0\)
\(\displaystyle
\frac{x^4 - 3x^3 + 3x^2 - x}{x^4 - 2x^3 + 2x - 1}\)I got \(\displaystyle \frac{0}{6} = 0\)