Pardon my mistakes in asking t I have problems solving equations of LIMITS. To be specific the ones I have to rationalize. https://www.ck12.org/book/ck-12-precalculus-concepts/section/14.5/
Example 5.
The last example is what's giving me issues. The denominator at the $4$-th equation to be precise. I don't know how it was manipulated.
Am an undergraduate in the University of Calabar in Nigeria. Am studying for my exams which will commence on 22nd of this month. I believe the answer will help me pass maths exams with good grades and the answer will also help others who have same issue. And I'll definitely reffer my friends to this website. Admin please kindly approve as the responses will help me in no small way. Thanks
Evaluate the following limit:
$$\lim{x\to 0}\left(\frac{3}{x\sqrt{9−x}}-\frac1x\right)$$
\begin{align}
\lim_{x \to 0}\left(\frac{3}{x\sqrt{9−x}}-\frac1x\right) &= \lim_{x \to 0}\left(\frac{3}{x\sqrt{9−x}}-\frac{\sqrt{9-x}}{x\sqrt{9-x}}\right) \\
&=\lim_{x \to 0} \left(\frac{3-\sqrt{9-x}}{x\sqrt{9-x}} \right)\\
&= \lim_{x \to 0} \left( \frac{3-\sqrt{9-x}}{x\sqrt{9-x}}\cdot \frac{3+\sqrt{9-x}}{3+\sqrt{9-x}}\right)\\
&= \lim_{x \to 0} \left( \frac{9-(9-x)}{x\sqrt{9-x}}\right)\\
&= \lim_{x \to 0} \frac{x}{x\sqrt{9-x}}\\
&= \lim_{x \to 0} \frac1{\sqrt{9-x}}\\
&= \frac1{\sqrt{9-0}}\\
&= \frac13
\end{align}

The answer is indeed wrong.
\begin{align}
\lim_{x \to 0} \left( \frac{(3-\sqrt{9-x}}{x\sqrt{9-x}} \cdot \frac{3+\sqrt{9-x}}{3+\sqrt{9-x}} \right) &= \lim_{x \to 0} \frac{1}{3+\sqrt{9-x}}\lim_{x \to 0}\frac{9-(9-x)}{x\sqrt{9-x}}\\
&= \frac16 \cdot \lim_{x \to 0}\frac{x}{x\sqrt{9-x}}\\
&=\frac16 \cdot \lim_{x \to 0}\frac1{\sqrt{9-x}}\\
&=\frac1{18}
\end{align}
