Find lim as x to 1 (x^5-3x^4+2x²+x-1)/(x^7-4x+3)

Find $\displaystyle\lim_{x\to1}\frac{x^{5}-3x^{4}+2x^{2}+x-1}{x^{7}-4x+3}$.

Step 1: The form

\[ 1-3+2+1-1 = 0, \qquad 1-4+3 = 0, \]

so the form is $\dfrac00$ and, by the factor theorem, $x-1$ divides both polynomials.

Step 2: Divide the numerator

Synthetic division of $1,-3,0,2,1,-1$ by $x-1$ gives the coefficients $1,-2,-2,0,1$ with remainder $0$: $x^{5}-3x^{4}+2x^{2}+x-1 = \bigl(x-1\bigr)\bigl(x^{4}-2x^{3}-2x^{2}+1\bigr)$.

Step 3: Divide the denominator

Synthetic division of $1,0,0,0,0,0,-4,3$ by $x-1$ gives $1,1,1,1,1,1,-3$ with remainder $0$: $x^{7}-4x+3 = \bigl(x-1\bigr)\bigl(x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x-3\bigr)$.

Step 4: Cancel and substitute

For $x\neq1$ the common factor cancels, and the two quotients no longer vanish at $1$: \[\lim_{x\to1}\frac{x^{4}-2x^{3}-2x^{2}+1}{x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x-3} = \frac{1-2-2+1}{6-3} = \frac{-2}{3}.\]

Step 5: Verification

\[\bigl(x-1\bigr)\bigl(x^{4}-2x^{3}-2x^{2}+1\bigr) = x^{5}-2x^{4}-2x^{3}+x-x^{4}+2x^{3}+2x^{2}-1 = x^{5}-3x^{4}+2x^{2}+x-1 \;\;\checkmark\]

\[\bigl(x-1\bigr)\bigl(x^{6}+\cdots+x-3\bigr) = \bigl(x^{7}+x^{6}+\cdots+x^{2}-3x\bigr)-\bigl(x^{6}+\cdots+x-3\bigr) = x^{7}-4x+3 \;\;\checkmark\]

\[\text{Numerically at } x = 1.001:\ \text{the quotient is } -0.66401 \approx -\tfrac23 \;\;\checkmark\]