Correct Answer - Option 3 : 0
Calculation:
L = \(\rm\lim_{x\to\infty}{x^2+3x^3+5x^5\over3x^2+ x^6}\)
Dividing by x6 in numerator and denominator
L = \(\rm\lim_{x\to\infty}{{1\over x^4}+{3\over x^3}+{5\over x}\over{3\over x^4}+ 1}\)
Let \(\rm 1\over x \) = a ⇒ \(\rm\lim_{x\to\infty}\) =\(\rm\lim_{a\to0}\)
L = \(\rm\lim_{a\to0}{{a^4}+{3a^3}+{5a}\over{3a^4}+ 1}\)
Putting the limits we get
L = \(0\over 1\) = 0
