Skip to content
7.2 Q.II · Q27

Q.lim⁡x→3[1x−3−9xx3−27]\displaystyle\lim_{x\to 3}\left[\frac{1}{x-3}-\frac{9x}{x^3-27}\right]

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
48% · 66/137 Questions
✓ Free question

Since x3−27=(x−3)(x2+3x+9)x^3-27=(x-3)(x^2+3x+9), combine: 1x−3−9x(x−3)(x2+3x+9)=(x2+3x+9)−9x(x−3)(x2+3x+9)=x2−6x+9(x−3)(x2+3x+9)=(x−3)2(x−3)(x2+3x+9)=x−3x2+3x+9\dfrac{1}{x-3}-\dfrac{9x}{(x-3)(x^2+3x+9)}=\dfrac{(x^2+3x+9)-9x}{(x-3)(x^2+3x+9)}=\dfrac{x^2-6x+9}{(x-3)(x^2+3x+9)}=\dfrac{(x-3)^2}{(x-3)(x^2+3x+9)}=\dfrac{x-3}{x^2+3x+9}. At x=3x=3: numerator is 00 and denominator is 2727, so the limit is 00.

✓Final answer

00

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.