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Limit x stands to 0(x^2 +3x-4) |
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Answer» -STEP explanation:= x→2lim x 4 −8x 2 +4 2 x 3 −3x 2 +4 Solve,Numeratorthen, put x=2,2 3 −3×2 2 +4=0Divide x−2)x 3 −3x 2 +4(x 2 −x−2 x 3 −2X 2 _______________ −x 2 +4 −x 2 +2x ______________ −2x+4 −2x+4 ____________ 0Now, x 3 −3x 2 +4=(x−2)(x 2 −x−2)x→2lim x 4 −8x 2 +4 2 x 3 −3x 2 +4 = x→2lim (x 2 −4 2 )(x−2)(x 2 −x−2) = x→2lim (x 2 −4)(x 2 −4)(x−2)(x 2 −(2−1)x−2) = x→2lim (x 2 −2 2 )(x 2 −2 2 )(x−2)(x 2 −2x+1x−2) = x→2lim (x−2)(x+2)(x−2)(x+2)(x−2)(x(x−2)+1(x−2)) = x→2lim x−2)(x−2)(x+2)(x+2)(x−2)(x−2)(x+1) = x→2lim (x+2)(x+2)x+1 Now, TAKING limit and we get= (2+2)(2+2)2+1 = 4×43 = 163 Hence, this is the answer. |
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