Correct Answer - Option 1 : 1
Concept:
L'Hospital's Rule:
If the limit becomes \({0\over 0}\) or \({\pm\infty\over \pm\infty}\), it is solved by differentiating numerator and denominator.
Calculation:
On putting the limits we get
\(\rm \lim_{x\rightarrow \infty}{x^4 + 3x^2 + 5\over x^4 +x^2 -6}\) = \(\infty\over\infty\)
Applying L'Hospital Rule
L(say) = \(\rm \lim_{x\rightarrow \infty}{4x^3 + 6x\over 4x^3 +2x}\) = \(\rm \lim_{x\rightarrow \infty}{4x^2+ 6\over 4x^2 +2}\) = \(\infty\over \infty\)
Again applying L'Hospital Rule
⇒ L = \(\rm \lim_{x\rightarrow \infty}{8x\over 8x}\)
⇒ \(\rm \lim_{x\rightarrow \infty}1\)
∴ L = 1
