\(\rm \lim_{x\rightarrow \infty}{x^4 + 3x^2 + 5\over x^4 +x^2 -6}\)

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