Determine $$ \lim _{x \rightarrow 9} \frac{9-x}{3-\sqrt{x}} $$

Question

UpStudy AI · Accepted Answer

We start by substituting $$ x = 9 $$ into the expression:

$$
\frac{9 - 9}{3 - \sqrt{9}} = \frac{0}{0},
$$

which is an indeterminate form.

Next, we recognize that the numerator can be factored as a difference of squares. Notice that:

$$
9 - x = 3^2 - (\sqrt{x})^2 = (3 - \sqrt{x})(3 + \sqrt{x}).
$$

Thus, the original expression becomes:

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

For $$ x \neq 9 $$, we can cancel the common factor $$ 3 - \sqrt{x} $$:

$$
\frac{(3 - \sqrt{x})(3 + \sqrt{x})}{3 - \sqrt{x}} = 3 + \sqrt{x}.
$$

Now, taking the limit as $$ x \rightarrow 9 $$:

$$
\lim_{x \rightarrow 9} \left(3 + \sqrt{x}\right) = 3 + \sqrt{9} = 3 + 3 = 6.
$$

Hence, the limit is:

$$
\boxed{6}
$$
The limit is 6.

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