2 votes 2 votes What should be the value of a,b and c such that the function defined below is continuous at x=0 ? $f\left ( x \right )=\begin{Bmatrix} \left ( 1+ax \right )^{\frac{1}{x}} & x<0 & \\ b & x=0& \\ \frac{(x+c)^{\frac{1}{3}}-1}{x} &x>0 & \end{Bmatrix}$ Calculus calculus engineering-mathematics continuity + – ManojK asked Dec 19, 2016 ManojK 1.0k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply vijaycs commented Dec 20, 2016 reply Follow Share a = b = infinity, if c = any finite value other than 1?? If c = 1, then , b = 1/3 , a = ln (1/3). ?? 1 votes 1 votes ManojK commented Dec 20, 2016 reply Follow Share Yes right.Provide the solution by considering second case. 0 votes 0 votes vijaycs commented Dec 20, 2016 i edited by vijaycs Dec 20, 2016 reply Follow Share See.. Since it is continuous at x = 0, so f(0-) = f(0+) = f(0). ------- (X) Here f(0-) = Lim(x->0) { (1+ax)1/x } = ea . f(0) = b and f(0+) =Lim ( x->0) {(x + c)1/3 - 1}/ (x) Now if c = 1 then, f(0+) =Lim ( x->0) {(x + 1)1/3 - 1} /(x) => 0/0 using L hospital rule, we can solve above limit = 1/3 . Hence from eqn (X) , ea = b = 1/3. a = ln (1/3) b = 1/3 c = 1. Is it okay now ? 3 votes 3 votes ManojK commented Dec 20, 2016 reply Follow Share yes. 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes $a= $any value $b=1$ $c$=$x^{3}+2x+3x^{2}+1$ Then x will be continuous at x=1 But continuous in x=0 not possible here srestha answered Dec 20, 2016 srestha comment Share Follow See all 2 Comments See all 2 2 Comments reply ManojK commented Dec 20, 2016 reply Follow Share No .Actually we have given that the above function is continuous at x=0 so find the value of a,b,c. 0 votes 0 votes askeshavas commented Jan 19, 2019 reply Follow Share it's wrong ma'am 0 votes 0 votes Please log in or register to add a comment.