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7.3 Q.II · Q45

Q.lim⁡x→0[x2+9−2x2+93x2+4−2x2+4]\displaystyle\lim_{x\to 0}\left[\frac{\sqrt{x^2+9}-\sqrt{2x^2+9}}{\sqrt{3x^2+4}-\sqrt{2x^2+4}}\right]

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Multiply the numerator by its conjugate x2+9+2x2+9\sqrt{x^2+9}+\sqrt{2x^2+9}: it becomes (x2+9)−(2x2+9)=−x2(x^2+9)-(2x^2+9)=-x^2. Multiply the denominator by its conjugate 3x2+4+2x2+4\sqrt{3x^2+4}+\sqrt{2x^2+4}: it becomes (3x2+4)−(2x2+4)=x2(3x^2+4)-(2x^2+4)=x^2. So the expression becomes $\dfrac{-x^2/(\sqrt{x^2+9}+\sqrt{2x^2+9})}{x^2/(\sqrt{3x^2+4}+\sqrt{2x^2+4})}=\dfrac{-(\sqrt{3x^2+4}+\sqrt …

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