Skip to content
7.2 Q.III · Q32

Q.lim⁡y→1/2[1−8y3y−4y3]\displaystyle\lim_{y\to 1/2}\left[\frac{1-8y^3}{y-4y^3}\right]

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
52% · 71/137 Questions
✓ Free question

1−8y3=1−(2y)3=(1−2y)(1+2y+4y2)1-8y^3=1-(2y)^3=(1-2y)(1+2y+4y^2). y−4y3=y(1−4y2)=y(1−2y)(1+2y)y-4y^3=y(1-4y^2)=y(1-2y)(1+2y). Cancelling (1−2y)(1-2y): 1+2y+4y2y(1+2y)\dfrac{1+2y+4y^2}{y(1+2y)}. At y=12y=\tfrac12: numerator =1+1+1=3=1+1+1=3, denominator =12(1+1)=1=\tfrac12(1+1)=1. Limit =31=3=\dfrac31=3.

✓Final answer

33

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.