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Now what? \[\Large \mathop {\lim }\limits_{x \to 3} {\rm{ \;\;log}}\frac{{x - 3}}{{\sqrt {x + 6} - 3}}\]
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Do you know Calculus and L'Hôpital's Rule ? See http://en.wikipedia.org/wiki/L'Hôpital's_rule If you do, take the derivative of the numerator and denominator, and evaluate each at x=3 \( \frac{d}{dx}(x-3)= 1\) \( \frac{d}{dx} (x+6)^{\frac{1}{2}} -3= \frac{1}{2\sqrt{x+6}}\) and we have \[ \frac{1}{\frac{1}{2\sqrt{x+6}}}= 2\sqrt{x+6} \] As x \( \to \) 3 this approaches 6
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