3x3 + 5x2 + 7x + 4 lim x+ 2x3 + x2 - x - 2
Answer (1)
1st Oct, 2021
This is a sample answer, try the same method for the question ypu had asked.
x→2limx4−8x2+16x3−3x2+4=x→2limx4−8x2+42x3−3x2+4Solve,Numerator then, put x=2,23−3×22+4=0
Divide x−2)x3−3x2+4(x2−x−2 x3−2x2 _______________ −x2+4 −x2+2x ______________ −2x+4 −2x+4 ____________ 0
Now, x3−3x2+4=(x−2)(x2−x−2)x→2limx4−8x2+42x3−3x2+4=x→2lim(x2−42)(x−2)(x2−x−2)=x→2lim(x2−4)(x2−4)(x−2)(x2−(2−1)x−2)=x→2lim(x2−22)(x2−22)(x−2)(x2−2x+1x−2)=x→2lim(x−2)(x+2)(x−2)(x+2)(x−2)(x(x−2)+1(x−2))=x→2limx−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=163Hence, this is the answer.
Hope this will help you.
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