2 votes 2 votes limx->0 (cot x)1/logx ? ans: 0(m getting 1/e) Vivek sharma asked Jul 20, 2015 Vivek sharma 3.3k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Alright ! Getting 1/e only. Where have u seen the answer from ? saloni answered Jul 21, 2015 saloni comment Share Follow See 1 comment See all 1 1 comment reply Vivek sharma commented Jul 21, 2015 reply Follow Share from b.s grewal book 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes lt x->0 (cotx)^(1/logx) lny=lt x->0 log(cotx)/(logx) =lt x->0 (-cosecx^2 * x/cotx) using l-hospitals =lt x->0 -x/(sinx*cosx) =-1 y=0 Sudeep Choudhary answered Jul 21, 2015 Sudeep Choudhary comment Share Follow See all 2 Comments See all 2 2 Comments reply Vivek sharma commented Jul 22, 2015 reply Follow Share log y = -1 will give y = 1/e not 0. 1 votes 1 votes Kaluti commented Sep 2, 2017 reply Follow Share yes i am also getting (1/e) as answer here 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes $\lim_{x \to 0} \cot(x)^{\frac{1}{\ln(x)}} = \lim_{x \to 0} e^{\frac{1}{\ln(x)}\times \ln(\cot(x)) }\\ \lim_{x \to 0}\frac{\ln(\cot(x))}{\ln(x)} = -1\\ \lim_{u \to -1}e^u = e^-1 = \frac{1}{e}$ Salman answered Aug 5, 2015 Salman comment Share Follow See all 0 reply Please log in or register to add a comment.