Limx- căn2x 3 / 9-x^2

To evaluate the limit

\[
\lim_{x \to 3} \frac{x - \sqrt{2x + 3}}{9 - x^2},
\]

let's begin by substituting \(x = 3\) directly into the expression.

First, the numerator:

\[
3 - \sqrt{2(3) + 3} = 3 - \sqrt{6 + 3} = 3 - \sqrt{9} = 3 - 3 = 0.
\]

Now for the denominator:

\[
9 - 3^2 = 9 - 9 = 0.
\]

Since substituting \(x = 3\) gives both the numerator and denominator equal to zero, we can apply L'Hôpital's Rule, which is useful when we have a \(\frac{0}{0}\) form. L'Hôpital's Rule states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \(\frac{\pm \infty}{\pm \infty}\), then

\[
\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)},
\]

assuming the limit on the right exists.

Let's differentiate the numerator and the denominator.

1. **Numerator:**

\[
f(x) = x - \sqrt{2x + 3}.
\]

Using the chain rule:

\[
f'(x) = 1 - \frac{1}{2\sqrt{2x + 3}} \cdot 2 = 1 - \frac{1}{\sqrt{2x + 3}}.
\]

2. **Denominator:**

\[
g(x) = 9 - x^2.
\]

Differentiating gives:

\[
g'(x) = -2x.
\]

Applying L'Hôpital's Rule:

\[
\lim_{x \to 3} \frac{f'(x)}{g'(x)} = \lim_{x \to 3} \frac{1 - \frac{1}{\sqrt{2x + 3}}}{-2x}.
\]

Now we can substitute \(x = 3\) into the derivatives we derived:

For the numerator:

\[
f'(3) = 1 - \frac{1}{\sqrt{2(3) + 3}} = 1 - \frac{1}{\sqrt{9}} = 1 - \frac{1}{3} = \frac{2}{3}.
\]

For the denominator:

\[
g'(3) = -2(3) = -6.
\]

So we have:

\[
\lim_{x \to 3} \frac{f'(x)}{g'(x)} = \frac{\frac{2}{3}}{-6} = \frac{2}{3} \cdot -\frac{1}{6} = -\frac{2}{18} = -\frac{1}{9}.
\]

Therefore, the final answer is:

\[
\lim_{x \to 3} \frac{x - \sqrt{2x + 3}}{9 - x^2} = -\frac{1}{9}.
\]