GATE CS 2019 Q13 — Calculus: Limits, Continuity, Differentiability

Compute $\displaystyle \lim_{x \rightarrow 3} \frac{x^4-81}{2x^2-5x-3}$

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Answer: C

To find the limit:

L = \lim_{x \rightarrow 3} \frac{x^4-81}{2x^2-5x-3}

First, let's check the limit by direct substitution of $x = 3$ :

Since the limit is in the indeterminate form $\frac{0}{0}$ , we can evaluate it using either factorization or L'Hôpital's Rule.

Factorize both the numerator and the denominator:

Numerator: $x^4 - 81 = (x^2 - 9)(x^2 + 9) = (x - 3)(x + 3)(x^2 + 9)$
Denominator: $2x^2 - 5x - 3 = 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (x - 3)(2x + 1)$

Substitute these factors back into the limit expression:

L = \lim_{x \rightarrow 3} \frac{(x - 3)(x + 3)(x^2 + 9)}{(x - 3)(2x + 1)}

Since $x \neq 3$ , we can cancel the common factor $(x - 3)$ :

L = \lim_{x \rightarrow 3} \frac{(x + 3)(x^2 + 9)}{2x + 1}

Now, substitute $x = 3$ :

L = \frac{(3 + 3)(3^2 + 9)}{2(3) + 1} = \frac{6 \times 18}{7} = \frac{108}{7}

Differentiating the numerator and the denominator with respect to $x$ :

L = \lim_{x \rightarrow 3} \frac{\frac{d}{dx}(x^4 - 81)}{\frac{d}{dx}(2x^2 - 5x - 3)}

L = \lim_{x \rightarrow 3} \frac{4x^3}{4x - 5}

Substituting $x = 3$ :

L = \frac{4(3)^3}{4(3) - 5} = \frac{4(27)}{12 - 5} = \frac{108}{7}

Thus, the correct option is C.

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Published 10 Oct 2026

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