Question:
If the following equation holds,
\lim_{x \rightarrow 2}\frac{\sqrt{9+ax}+b}{x-2}=-2
Find a and b.
Answer:
Since the extreme value of the denominator becomes zero (\lim_{x\rightarrow 2}(x-2)=0), the numerator must also be zero.
\lim_{x \rightarrow 2}(\sqrt{9+ax}+b)=\sqrt{9+2a}+b=0 \\
\rightarrow b = -\sqrt{9+2a}
Then, plug b into the original equation and rationalize the numerator and denominator.
\begin{eqnarray*}
\lim_{x \rightarrow 2}\frac{\sqrt{9+ax}-\sqrt{9+2a}}{x-2}
&=& \lim_{x \rightarrow 2}\frac{\sqrt{9+ax}-\sqrt{9+2a}}{x-2}\times \frac{\sqrt{9+ax}+\sqrt{9+2a}}{\sqrt{9+ax}+\sqrt{9+2a}} \\
&=& \lim_{x \rightarrow 2}\frac{(9+ax)-(9+2a)}{(x-2)(\sqrt{9+ax}+\sqrt{9+2a})} \\
&=& \lim_{x \rightarrow 2}\frac{a(x-2)}{(x-2)(\sqrt{9+ax}+\sqrt{9+2a})} \\
&=& \frac{a}{2\sqrt{9+2a}}
\end{eqnarray*}
Now, this result is equal to -2. Thus,
\frac{a}{2\sqrt{9+2a}} = -2 \\
\rightarrow a = -4\sqrt{9+2a} \\
\rightarrow a^2 = 16(9+2a) \\
\rightarrow a^2-32a-144 = (a-36)(a+4)=0
The extreme value a = -4\sqrt{9+2a} obtained above indicates that a<0. Therefore, a=-4. We know b = -\sqrt{9+2a}, so b=-1.
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