Skip to content
7.3 Q.I · Q37

Q.lim⁡x→0[6+x+x2−6x]\displaystyle\lim_{x\to 0}\left[\frac{\sqrt{6+x+x^2}-\sqrt6}{x}\right]

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
55% · 76/137 Questions
✓ Free question

Multiply top and bottom by 6+x+x2+6\sqrt{6+x+x^2}+\sqrt6: the numerator becomes (6+x+x2)−6=x+x2=x(1+x)(6+x+x^2)-6=x+x^2=x(1+x). Dividing by the outer xx: 1+x6+x+x2+6\dfrac{1+x}{\sqrt{6+x+x^2}+\sqrt6}. At x=0x=0: 16+6=126=612\dfrac{1}{\sqrt6+\sqrt6}=\dfrac{1}{2\sqrt6}=\dfrac{\sqrt6}{12}.

✓Final answer

612\dfrac{\sqrt6}{12}

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.