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7.3 Q.II · Q42

Q.lim⁡x→2[x2−4x+2−3x−2]\displaystyle\lim_{x\to 2}\left[\frac{x^2-4}{\sqrt{x+2}-\sqrt{3x-2}}\right]

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Multiply by the conjugate x+2+3x−2\sqrt{x+2}+\sqrt{3x-2}: the denominator becomes (x+2)−(3x−2)=4−2x=−2(x−2)(x+2)-(3x-2)=4-2x=-2(x-2). So the expression is (x2−4)(x+2+3x−2)−2(x−2)=(x−2)(x+2)(x+2+3x−2)−2(x−2)=(x+2)(x+2+3x−2)−2\dfrac{(x^2-4)(\sqrt{x+2}+\sqrt{3x-2})}{-2(x-2)}=\dfrac{(x-2)(x+2)(\sqrt{x+2}+\sqrt{3x-2})}{-2(x-2)}=\dfrac{(x+2)(\sqrt{x+2}+\sqrt{3x-2})}{-2} after cancelling (x−2)(x-2). At x=2x=2: 4(2+2)−2=16−2=−8\dfrac{4(2+2)}{-2}=\dfrac{16}{-2}=-8.

✓Final answer

−8-8

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