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7.3 Q.I · Q38

Q.lim⁡x→3[2x+3−4x−3x2−9]\displaystyle\lim_{x\to 3}\left[\frac{\sqrt{2x+3}-\sqrt{4x-3}}{x^2-9}\right]

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✓ Free question

Multiply by the conjugate 2x+3+4x−3\sqrt{2x+3}+\sqrt{4x-3}: numerator becomes (2x+3)−(4x−3)=6−2x=−2(x−3)(2x+3)-(4x-3)=6-2x=-2(x-3). So the expression is −2(x−3)(x2−9)(2x+3+4x−3)=−2(x−3)(x−3)(x+3)(2x+3+4x−3)=−2(x+3)(2x+3+4x−3)\dfrac{-2(x-3)}{(x^2-9)(\sqrt{2x+3}+\sqrt{4x-3})}=\dfrac{-2(x-3)}{(x-3)(x+3)(\sqrt{2x+3}+\sqrt{4x-3})}=\dfrac{-2}{(x+3)(\sqrt{2x+3}+\sqrt{4x-3})} after cancelling (x−3)(x-3). At x=3x=3: −26×(3+3)=−236=−118\dfrac{-2}{6\times(3+3)}=\dfrac{-2}{36}=-\dfrac{1}{18}.

✓Final answer

−118-\dfrac{1}{18}

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