Evaluate `lim_(xtosqrt(10)) (sqrt(7+2x)-(sqrt(5)+sqrt(2)))/(x^(2)-10).`

Explanation

To evaluate the limit

\[ \lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - (\sqrt{5} + \sqrt{2})}{x^2 - 10}, \]

we will follow these steps:

### Step 1: Substitute \(x = \sqrt{10}\)

First, we substitute \(x = \sqrt{10}\) into the expression:

\[ \sqrt{7 + 2\sqrt{10}} - (\sqrt{5} + \sqrt{2}) \quad \text{and} \quad (\sqrt{10})^2 - 10. \]

Calculating the denominator:

\[ (\sqrt{10})^2 - 10 = 10 - 10 = 0. \]

Now calculating the numerator:

\[ \sqrt{7 + 2\sqrt{10}} - (\sqrt{5} + \sqrt{2}). \]

We need to check if the numerator also approaches zero:

\[ \sqrt{7 + 2\sqrt{10}} = \sqrt{5} + \sqrt{2}. \]

Thus, substituting \(x = \sqrt{10}\) gives us the form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule

Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator.

#### Derivative of the Numerator:

Let \( f(x) = \sqrt{7 + 2x} - (\sqrt{5} + \sqrt{2}) \).

The derivative \( f'(x) \) is:

\[ f'(x) = \frac{1}{2\sqrt{7 + 2x}} \cdot 2 = \frac{1}{\sqrt{7 + 2x}}. \]

#### Derivative of the Denominator:

Let \( g(x) = x^2 - 10 \).

The derivative \( g'(x) \) is:

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

### Step 3: Rewrite the Limit

Now we rewrite the limit using the derivatives:

\[ \lim_{x \to \sqrt{10}} \frac{f'(x)}{g'(x)} = \lim_{x \to \sqrt{10}} \frac{\frac{1}{\sqrt{7 + 2x}}}{2x}. \]

### Step 4: Substitute \(x = \sqrt{10}\) Again

Now we substitute \(x = \sqrt{10}\) into the new limit expression:

\[ = \frac{1}{\sqrt{7 + 2\sqrt{10}} \cdot 2\sqrt{10}}. \]

Calculating \( \sqrt{7 + 2\sqrt{10}} \):

\[ \sqrt{7 + 2\sqrt{10}} = \sqrt{5} + \sqrt{2}. \]

Thus, we have:

\[ = \frac{1}{(\sqrt{5} + \sqrt{2}) \cdot 2\sqrt{10}}. \]

### Step 5: Final Result

The final result of the limit is:

\[ \lim_{x \to \sqrt{10}} \frac{\sqrt{7 + 2x} - (\sqrt{5} + \sqrt{2})}{x^2 - 10} = \frac{1}{(\sqrt{5} + \sqrt{2}) \cdot 2\sqrt{10}}. \]