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7.3 Q.III · Q48

Q.lim⁡x→4[x2+x−203x+4−4]\displaystyle\lim_{x\to 4}\left[\frac{x^2+x-20}{\sqrt{3x+4}-4}\right]

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Numerator factors as (x−4)(x+5)(x-4)(x+5). Rationalizing the denominator by multiplying with 3x+4+4\sqrt{3x+4}+4: (3x+4)−16=3x−12=3(x−4)(3x+4)-16=3x-12=3(x-4). So the expression is (x−4)(x+5)(3x+4+4)3(x−4)=(x+5)(3x+4+4)3\dfrac{(x-4)(x+5)(\sqrt{3x+4}+4)}{3(x-4)}=\dfrac{(x+5)(\sqrt{3x+4}+4)}{3} after …

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