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Miscellaneous 7 I · Q99

Q.Select the correct answer from the given alternatives. lim⁡x→2(x4−16x2−5x+6)=\displaystyle\lim_{x\to 2}\left(\frac{x^4-16}{x^2-5x+6}\right)= (A) 23 (B) 32 (C) −32-32 (D) −16-16

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x4−16=(x2−4)(x2+4)=(x−2)(x+2)(x2+4)x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4); x2−5x+6=(x−2)(x−3)x^2-5x+6=(x-2)(x-3). Cancelling (x−2)(x-2): (x+2)(x2+4)x−3\dfrac{(x+2)(x^2+4)}{x-3}. At x=2x=2: 4×8−1=−32\dfrac{4\times8}{-1}=-32.

✓Final answer

−32-32 (option C)

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