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7.3 Q.III · Q50

Q.lim⁡x→0(3x9−x−1x)\displaystyle\lim_{x\to 0}\left(\frac{3}{x\sqrt{9-x}}-\frac{1}{x}\right)

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Write 3x9−x−1x=3−9−xx9−x\dfrac{3}{x\sqrt{9-x}}-\dfrac1x=\dfrac{3-\sqrt{9-x}}{x\sqrt{9-x}}. Rationalizing the numerator by multiplying with 3+9−x3+\sqrt{9-x}: it becomes 9−(9−x)=x9-(9-x)=x. So the expression becomes $\dfrac{x}{x\sqrt{9-x}\left(3+\sqrt{9-x}\right)}=\dfrac{1}{\sqrt{9-x}\left(3+\sqr …

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