Skip to content
NCERT Exemplar · Q13

Q.Evaluate lim⁡x→12(8x−32x−1−4x2+14x2−1)\lim_{x \to \frac{1}{2}} \left( \dfrac{8x - 3}{2x - 1} - \dfrac{4x^2 + 1}{4x^2 - 1} \right).

Uttar Pradesh UpmspShort· 3mImportance★★★★★est
62% · 108/175 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Each fraction blows up on its own at x=12x=\frac12, so we combine them over a common denominator, factor, cancel the (2x−1)(2x-1) that causes the 00\frac{0}{0}, and then substitute. The limit is 72\dfrac{7}{2}.

Step 1 — See why we cannot substitute directly

At x=12x = \frac{1}{2} both denominators vanish: 2x−1=02x-1 = 0 and 4x2−1=04x^2 - 1 = 0. So each fraction alone is undefined; we must combine them first.

Step 2 — Factor and take a common denominator

Note 4x2−1=(2x−1)(2x+1)4x^2 - 1 = (2x-1)(2x+1), so the common denominator is (2x−1)(2x+1)(2x-1)(2x+1):

8x−32x−1−4x2+1(2x−1)(2x+1)=(8x−3)(2x+1)−(4x2+1)(2x−1)(2x+1).\frac{8x-3}{2x-1} - \frac{4x^2+1}{(2x-1)(2x+1)} = \frac{(8x-3)(2x+1) - (4x^2+1)}{(2x-1)(2x+1)}.

Step 3 — Simplify the numerator

(8x−3)(2x+1)=16x2+2x−3,(8x-3)(2x+1) = 16x^2 + 2x - 3,

so the numerator is

16x2+2x−3−(4x2+1)=12x2+2x−4=2(6x2+x−2).16x^2 + 2x - 3 - (4x^2 + 1) = 12x^2 + 2x - 4 = 2(6x^2 + x - 2).

Factor the quadratic: 6x2+x−2=(2x−1)(3x+2)6x^2 + x - 2 = (2x-1)(3x+2). Hence …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.