Extreme value 1

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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