Since the limit becomes the undetermined form
[tex] \displaystyle \lim_{x\to 1} \dfrac{x^3-2x^2+3x-2}{2x^4-3x+1} \to \dfrac{0}{0} [/tex]
it means that both polynomials have a root at [tex] x=1 [/tex]. So, we can fact both numerator and denominator:
[tex] x^3-2x^2+3x-2 = (x-1)(x^2-x+2) [/tex]
[tex] 2x^4-3x+1 = (x-1)(2x^3+2x^2+2x-1) [/tex]
So, the fraction becomes
[tex] \dfrac{(x-1)(x^2-x+2)}{(x-1)(2x^3+2x^2+2x-1)} = \dfrac{x^2-x+2}{2x^3+2x^2+2x-1} [/tex]
Now, as x approaches 1, you have no problems anymore:
[tex] \displaystyle \lim_{x\to 1} \dfrac{x^2-x+2}{2x^3+2x^2+2x-1} \to \dfrac{2}{5} [/tex]
