0 votes 0 votes How to solve this? Calculus engineering-mathematics calculus limits virtual-gate-test-series + – Purple asked Feb 1, 2016 edited Apr 2, 2019 by Lakshman Bhaiya Purple 733 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Digvijay Pandey commented Feb 1, 2016 reply Follow Share Actually der is typo in question.. question was like limit x----> -1 0 votes 0 votes Purple commented Feb 2, 2016 reply Follow Share then we can solve by LH rule right? by converting $\frac{ln(x+1))}{\frac{1}{^{(x+1)^1/3}}}$ 0 votes 0 votes Please log in or register to add a comment.
Best answer 3 votes 3 votes option B now i rectifed the typo and solved i.e x-> 1- to x->-1 let x+1=z now when x-> -1 then z->0 so we can re write it in this way lim z->0 ln z /z^-3 its an indeterminate form we can apply LH rule lim z->0 1/z * 3z^2 lim z->0 3z =0 Bhagirathi answered Feb 4, 2016 edited Dec 20, 2022 by Sachin Mittal 1 Bhagirathi comment Share Follow See 1 comment See all 1 1 comment reply __ commented Sep 5, 2016 reply Follow Share as far as i understood its z=1+x lt z->0 z^(1/3) * lnz on Lhospital rule lt Z=$lim z->0 ln z/(z^-1/3)) =limz->0 z^-1/3 /(-1/3) =0$ 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes if we take floor value itz =0 if ceiling value itz=1 Tauhin Gangwar answered Feb 1, 2016 Tauhin Gangwar comment Share Follow See 1 comment See all 1 1 comment reply Purple commented Feb 1, 2016 reply Follow Share How to solveee?? step wise plz 0 votes 0 votes Please log in or register to add a comment.