5.4. For what value of $$ \kappa $$ is the function continuous? $$ g(x)=\left\{\begin{array}{ll} \frac{\sqrt{7 x+2}-\sqrt{6 x+4}}{x-2} & \text { if } x \geq-\frac{2}{7} \text { and } x \neq 2 \ \kappa & \text { if } x=2 \end{array}\right. $$

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To ensure continuity at $$ x=2 $$, we must have
$$
\lim_{x \to 2} g(x)=g(2)=\kappa.
$$

Since
$$
g(x)=\frac{\sqrt{7x+2}-\sqrt{6x+4}}{x-2} \quad \text{for } x \neq 2,
$$
we begin by evaluating the limit:
$$
\lim_{x \to 2} \frac{\sqrt{7x+2}-\sqrt{6x+4}}{x-2}.
$$

**Step 1: Multiply by the Conjugate**

Multiply the numerator and denominator by the conjugate $$ \sqrt{7x+2}+\sqrt{6x+4} $$:
$$
\frac{\sqrt{7x+2}-\sqrt{6x+4}}{x-2}
\cdot \frac{\sqrt{7x+2}+\sqrt{6x+4}}{\sqrt{7x+2}+\sqrt{6x+4}}.
$$

**Step 2: Simplify the Expression**

The numerator becomes:
$$
(\sqrt{7x+2}-\sqrt{6x+4})(\sqrt{7x+2}+\sqrt{6x+4}) = (7x+2) - (6x+4).
$$
Simplify the numerator:
$$
(7x+2) - (6x+4) = x - 2.
$$
Thus, the expression simplifies to:
$$
\frac{x-2}{(x-2)(\sqrt{7x+2}+\sqrt{6x+4})}.
$$
For $$ x \neq 2 $$, cancel the $$ x-2 $$ terms:
$$
\frac{1}{\sqrt{7x+2}+\sqrt{6x+4}}.
$$

**Step 3: Evaluate the Limit as $$ x \to 2 $$**

Substitute $$ x=2 $$:
$$
\sqrt{7(2)+2} = \sqrt{14+2} = \sqrt{16} = 4,
$$
$$
\sqrt{6(2)+4} = \sqrt{12+4} = \sqrt{16} = 4.
$$
Thus,
$$
\lim_{x \to 2}\frac{1}{\sqrt{7x+2}+\sqrt{6x+4}} = \frac{1}{4+4} = \frac{1}{8}.
$$

Since continuity at $$ x=2 $$ requires $$ g(2)=\kappa $$ to be equal to the limit, we have:
$$
\kappa = \frac{1}{8}.
$$
$$ \kappa = \frac{1}{8} $$

5.4. For what value of is the function continuous?

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