Question

lim(x→9) (9 - x)/(3 - √x)

Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator $3 + \sqrt{x}$.
\[

$$\begin{align*} \lim_{x ightarrow9}\frac{9 - x}{3-\sqrt{x}}&=\lim_{x ightarrow9}\frac{(9 - x)(3+\sqrt{x})}{(3-\sqrt{x})(3+\sqrt{x})}\\ \end{align*}$$

\]

Step2: Simplify the denominator

Use the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$. Here $a = 3$ and $b=\sqrt{x}$, so $(3-\sqrt{x})(3+\sqrt{x})=9 - x$.
\[

$$\begin{align*} \lim_{x ightarrow9}\frac{(9 - x)(3+\sqrt{x})}{(3-\sqrt{x})(3+\sqrt{x})}&=\lim_{x ightarrow9}\frac{(9 - x)(3+\sqrt{x})}{9 - x}\\ \end{align*}$$

\]

Step3: Cancel out the common factor

Cancel out the common factor $(9 - x)$ (since $x
eq9$ when taking the limit).
\[

$$\begin{align*} \lim_{x ightarrow9}\frac{(9 - x)(3+\sqrt{x})}{9 - x}&=\lim_{x ightarrow9}(3+\sqrt{x}) \end{align*}$$

\]

Step4: Evaluate the limit

Substitute $x = 9$ into $3+\sqrt{x}$.
\[

$$\begin{align*} \lim_{x ightarrow9}(3+\sqrt{x})&=3+\sqrt{9}\\ &=3 + 3\\ &=6 \end{align*}$$

\]

Answer:

$6$