Correct Answer - Option 1 : 1
Concept:
L'Hospital's Rule:
If the limit becomes \({0\over 0}\) or \({\pm∞\over \pm∞}\), it is solved by differentiating numerator and denominator.
Calculation:
Let L = \(\rm\lim_{x\rightarrow ∞}{x^3+3x^2+6x+5\over x^3+2x+6}\)
Putting x = ∞, we get L = \({\pm∞\over \pm∞}\)
Differentiating numerator and denominator
L = \(\rm\lim_{x\rightarrow ∞}{3x^2+6x+6\over 3x^2+2}\)
Again putting x = ∞, we get L = \({\pm∞\over \pm∞}\)
Differentiating numerator and denominator
L = \(\rm\lim_{x\rightarrow ∞}{6x+6\over 6x}\)
Again putting x = ∞, we get L = \({\pm∞\over \pm∞}\)
Differentiating numerator and denominator
L = \(\rm\lim_{x\rightarrow ∞}{6\over 6}\) = \(\rm\lim_{x\rightarrow ∞}1\)
∴ L = 1
