Given x→0lim3x3sin−1x−tan−1x
⇒x→0lim3x3(x+3!x3…)−(x−3x3…)
⇒x→0lim9x2(1+3!3x2…)−(1−33x2…) (Use L'Hospital rule 00 form)
⇒x→0lim18x(3!6x+…)−(−36x+…) (Use L'Hospital rule 00 form)
⇒x→0lim18(3!6+…)−(−36+…)=181+2=61
Given that, x→0lim3x3sin−1x−tan−1x=L
∴61=L⇒6L=1
So, 6L+1=2