Question

# Find the limits : i. limx→1[x2+1x+100] ii. limx→2[x3−4x2+4xx2−4] iii. limx→2[x2−4x3−4x2+4x] iv. limx→2[x3−2x2x2−5x+6] v. limx→1[x−2x2−x−1x3−3x2+2x]

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Solution

## (i) Given : limx→1[x2+1x+100] Substituting x=1 =12+11+100 =1+1101 =2101 ∴limx→1[x2+1x+100]=2101 (ii) Given : limx→2[x3−4x2+4xx2−4] Substituting the given value, limx→2[x3−4x2+4xx2−4]=23−4×22+4×222−4 =8−16+84−4 =00 Since it is in 00 form, we have to simplify it. ⇒limx→2[x3−4x2+4xx2−4] =limx→2x(x2−4x+4)x2−(2)2 =limx→2x(x2+22−2⋅2⋅x)(x−2)(x+2) =limx→2x(x−2)2(x−2)(x+2) (∵x≠2) =limx→2x(x−2)x+2 Substituing x=2 =2(2−2)2+2=2(0)4=04=0 (iii) Given : limx→2[x2−4x3−4x2+4x] Substituting the given value, limx→2[x2−4x3−4x2+4x]=22−423−4×22+4×2 =4−48−16+8 =00 Since it is in 00 form, we have to simplify it. ⇒limx→2[x2−4x3−4x2+4x] =limx→2[x2−(2)2x(x2−4x+4)] =limx→2[(x−2)(x+2)x(x2+22−2⋅2⋅x)] =limx→2[(x−2)(x+2)x(x−2)2] =limx→2[x+2x(x−2)] (∵x≠2) Substituing x=2 =2+22(2−2)=42(0)=40 =∞ Which is not defined. (iv) Given : limx→2[x3−2x2x2−5x+6] Substituting the given value, limx→2[x3−2x2x2−5x+6]=23−2×2222−5×2+6 =8−84−10+6 =00 Since it is in 00 form, we have to simplify it. ⇒limx→2[x3−2x2x2−5x+6] =limx→2x2(x−2)x2−3x−2x+6 =limx→2(x2(x−2)x(x−3)−2(x−3)) =limx→2(x2(x−2)(x−2)(x−3)) =limx→2(x2x−3) (∵x≠2) Substituting x=2 =(2)22−3 =4−1 =−4 ∴limx→2[x3−2x2x2−5x+6]=−4 (v) Given : limx→1[x−2x2−x−1x3−3x2+2x] =limx→1[x−2x(x−1)−1x(x2−3x+2)] =limx→1[x−2x(x−1)−1x(x(x−2)−1(x−2))] =limx→1[x−2x(x−1)−1x(x−1)(x−2)] =limx→1[(x−2)(x−2)−1x(x−1)(x−2)] =limx→1[(x−2)2−12x(x−1)(x−2)] =limx→1[(x−2−1)(x−2+1)x(x−1)(x−2)] =limx→1[(x−3)(x−1)x(x−1)(x−2)] (∵x≠1) =limx→1[x−3x(x−2)] Substituing x=1 =1−31(1−2) =−21×−1 =2 ∴limx→1[x−2x2−x−1x3−3x2+2x]=2

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