# Limits, continuity and differentiability: GATE CS question with solution

From the GATE Daily set of 26 October 2026, question 2 of 10 (https://gatedaily.in/q/2026-10-26-limit-double-root-rational-function/). Difficulty: standard.

### Q.2 · Engineering Mathematics · Limits, continuity and differentiability · MCQ, 1 mark (choose one option)

The value of

$$\lim_{x\to 1}\frac{x^3-3x+2}{x^4-2x^2+1}$$

is

(A) $0$
(B) $\dfrac{3}{4}$
(C) $\dfrac{3}{2}$
(D) The limit does not exist

**Answer:** (B) $\dfrac{3}{4}$

**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):

$$\lim_{x\to1}\frac{x+2}{(x+1)^2} = \frac{3}{4}.$$

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

**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.

**Remember:** After each L'Hopital step or cancellation, substitute again: if the form is still 0/0, the root is repeated and you must go one more step.

**Why each option:**
- (A) Tempts if you notice only that the numerator vanishes at 1 and forget the denominator vanishes too.
- (B) Both have a double root at 1; cancelling $(x-1)^2$ leaves $(x+2)/(x+1)^2 = 3/4$.
- (C) Comes from factoring the denominator as $(x-1)^2(x+1)$ instead of $(x-1)^2(x+1)^2$.
- (D) Tempts after one L'Hopital step still gives 0/0; the form needs another step, not a verdict.