Lim X-> 1 (SQRT(x+3) - 2) / (x - 1) This Is What

Anonymous

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Lim x-> 1 (SQRT(x+3) - 2) / (x - 1) This is what I get (SQRT(x+3) - 2) (SQRT(x+3)+ 2) ---------------- *---------------- (1 - x)(SQRT(x+3)+ 2) x-1 ----------------------- (1 - x)(SQRT(x+3)+ 2) what did I do wrong or what didn't I finish? Thanks for the help!

 

Answer No.1

UberMonkey7

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Lim x-> 1 (SQRT(x+3) - 2) / (x - 1) = lim x-> 1 (SQRT(x+3) - 2) /(SQRT(x+3) - 2)(SQRT(x+3) + 2) =lim x-> 1 1/(SQRT(x+3) + 2) =1/4

Answer No.2

Anonymous

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Lim{x-->3} (sqrt(x+1) - 2) / (x-3) = lim{x-->3} (sqrt(x+1) - 2)(sqrt(x+1) + 2) / (x-3)(sqrt(x+1) + 2) = lim{x-->3} (x+1-4) / (x-3)(sqrt(x+1) + 2) = lim{x-->3} 1 / (sqrt(x+1) + 2) = 1/4

Answer No.3

Anonymous

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(sqrt(6 - x) - 2)/(sqrt(3 - x) - 1) Try multiplying top and bottom by (sqrt(3 - x) + 1): = ((sqrt(3 - x) + 1)*(sqrt(6 - x) - 2))/((sqrt(3 - x) - 1)*(sqrt(3 - x) + 1) Now multiply out: = (sqrt(6 - x)sqrt(3 - x) + sqrt(6 - x) + sqrt(3 - x) - 2)/(sqrt(3 - x)(sqrt(3 - x) - sqrt(3 - x) + sqrt(3 - x) - 1) Note that on the bottom, sqrt(3 - x)sqrtr(3 - x) = sqrt((3 - x)^2) = 3 - x, and two of the terms cancel, giving = (sqrt(6 - x)sqrt(3 - x) + sqrt(6 - x) + sqrt(3 - x) - 2)/(3 - x- 1) = (sqrt(6 - x)sqrt(3 - x) + sqrt(6 - x) + sqrt(3 - x) - 2)/(2 - x) Note that the bottom of the fraction is 2 - x, which is 0 at x = 2; no amount of rearranging will solve this as a singularity is a singularity however you express the fraction. Hence, you must use l'Hopital. Incidentally, the solution with l'Hopital is lim(x -> 2) [(sqrt(6 - x) - 2)/(sqrt(3 - x) - 1)] = lim(x -> 2) [((6 - x)^(1/2) - 2)/((3 - x)^(1/2) - 1)] d/dx[(6 - x)^(1/2) - 2] = (-1/2)(6 - x)^(-1/2): at x = 2, this equals (-1/2)(4)^(-1/2) = (-1/2)*(1/2) = -1/4. d/dx[((3 - x)^(1/2) - 1)] = (-1/2)(3 - x)^(-1/2): at x = 2, this equals (-1/2)(1)^(-1/2) = (-1/2). So the limit is (-1/4)/(-1/2) = 1/2.

Answer No.4

Anonymous

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Using L-hospital's rule that is differentiating numerator and denominator we get (1/sqrt[(x+3)-2])/1 applying limit we get 1/sqrt(2)

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