Question

calculate the following limit: \\(\lim\limits_{x\to 9} \dfrac{9 - x}{3 - \sqrt{x}}\\) \\(\bigcirc\\) -6 \\(\bigcirc\\) 0 \\(\bigcirc\\) 1 \\(\bigcirc\\) does not exist. \\(\bigcirc\\) 6

Explanation:

Step1: Factor the numerator

Notice that \(9 - x\) is a difference of squares, so we can factor it as \((3 - \sqrt{x})(3 + \sqrt{x})\) since \(9 - x = 3^2 - (\sqrt{x})^2=(3 - \sqrt{x})(3 + \sqrt{x})\). So the expression becomes:

$$\lim_{x ightarrow9}\frac{(3 - \sqrt{x})(3 + \sqrt{x})}{3 - \sqrt{x}}$$

Step2: Cancel common factors

We can cancel out the common factor \((3 - \sqrt{x})\) from the numerator and the denominator (assuming \(3 - \sqrt{x}
eq0\), which is true as \(x
ightarrow9\) but \(x
eq9\) in the limit process). After canceling, we get:

$$\lim_{x ightarrow9}(3 + \sqrt{x})$$

Step3: Evaluate the limit

Now we can directly substitute \(x = 9\) into the expression \(3+\sqrt{x}\) because the function \(3+\sqrt{x}\) is continuous at \(x = 9\). Substituting \(x = 9\), we have \(3+\sqrt{9}=3 + 3=6\).

Answer:

6