How do I evaluate this limit? lim(x->3) (x^2+2x-|x-3|)/(|x|+|x-3|+|x^2-9| Im con

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5 lim_(x->3) (-abs(x-3)+x^2+2 x)/(abs(x^2-9)+abs(x-3)+abs(x)) The limit of a quotient is the quotient of the limits: = (lim_(x->3) (x^2+2 x-abs(x-3)))/(lim_(x->3) (abs(x-3)+abs(x)+abs(x^2-9))) The limit of a sum is the sum of the limits: = (-lim_(x->3) abs(x-3)+lim_(x->3) x^2+2 (lim_(x->3) x))/(lim_(x->3) (abs(x-3)+abs(x)+abs(x^2-9))) The limit of x as x approaches 3 is 3: = (-lim_(x->3) abs(x-3)+lim_(x->3) x^2+6)/(lim_(x->3) (abs(x-3)+abs(x)+abs(x^2-9))) Using the power law, write lim_(x->3) x^2 as (lim_(x->3) x)^2: = (-lim_(x->3) abs(x-3)+(lim_(x->3) x)^2+6)/(lim_(x->3) (abs(x-3)+abs(x)+abs(x^2-9))) The limit of x as x approaches 3 is 3: = (15-lim_(x->3) abs(x-3))/(lim_(x->3) (abs(x-3)+abs(x)+abs(x^2-9))) Using the continuity of abs(x) at x = 0 write lim_(x->3) abs(x-3) as abs(lim_(x->3) (x-3)): = (15-abs(lim_(x->3) (x-3)))/(lim_(x->3) (abs(x-3)+abs(x)+abs(x^2-9))) The limit of x-3 as x approaches 3 is 0: = 15/(lim_(x->3) (abs(x-3)+abs(x)+abs(x^2-9))) The limit of a sum is the sum of the limits: = 15/(lim_(x->3) abs(x^2-9)+lim_(x->3) abs(x-3)+lim_(x->3) abs(x)) Using the continuity of abs(x) at x = 0 write lim_(x->3) abs(x-3) as abs(lim_(x->3) (x-3)): = 15/(lim_(x->3) abs(x^2-9)+abs(lim_(x->3) (x-3))+lim_(x->3) abs(x)) The limit of x-3 as x approaches 3 is 0: = 15/(lim_(x->3) abs(x^2-9)+lim_(x->3) abs(x)) The limit of abs(x) as x approaches 3 is 3: = 15/(lim_(x->3) abs(x^2-9)+3) Using the continuity of abs(x) at x = 0 write lim_(x->3) abs(x^2-9) as abs(lim_(x->3) (x^2-9)): = 15/(abs(lim_(x->3) (x^2-9))+3) The limit of x^2-9 as x approaches 3 is 0: = 5

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