AP EAMCET2006MathematicsLimits
If l₁= _ x 2⁺ (x+[x]), l₂ _ x 2⁻ (2 x-[x]) and l₃= _ x / 2 x (x- / 2) , then :
Options
- Al₁ < l₂ < l₃
- Bl₂ < l₃ < l₁
- Cl₃ < l₂ < l₁
- Dl₁ < l₃ < l₂
Correct answer
C. l₃ < l₂ < l₁
Step-by-step solution
l₁= _ x 2⁺ x+[x]= _ h 0 2+h+[2+h]=4l₂= _ x 2⁻ (2 x-[x])= _ h 0 2(2-h)-[2-h] = _ h 0 2(2-h)-1 =4-1=3l₃= _ x 2 x x- 2 = _ x 2 x=1 (By L' Hospital's rule) Thus, l₃ < l₂ < l₁