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.

Thank you