Limits, continuity and differentiability: GATE CS question with solution

Method: confirm 0/0, factor out the common root, substitute.

At $x=1$: numerator $1-3+2=0$, denominator $1-2+1=0$. So $x-1$ divides both.

• Numerator: $x^3-3x+2=(x-1)(x^2+x-2)=(x-1{)}^2(x+2)$.
• Denominator: $x^4-2x^2+1=(x^2-1{)}^2=(x-1{)}^2(x+1{)}^2$.

Cancel $(x-1{)}^2$ (allowed because $x\ne 1$ in a limit):

Same answer by L'Hopital. One pass gives $\frac{3x^2-3}{4x^3-4x}$, which is still $0/0$ at $x=1$. A second pass gives $\frac{6x}{12x^2-4}\to \frac{6}{8}=\frac{3}{4}$.

Trap. Both polynomials have a double root at 1, so a single L'Hopital pass (or a single factor of $x-1$) does not remove the indeterminacy. Stopping there leads to "does not exist" or 0; mis-factoring the denominator gives $3/2$.

What was tested: recognising a repeated root and re-checking the form after each simplification.