Let $g(x) = \frac{x-5}{\sqrt{x-4}-1}$ when $x \neq 5$ . $g$ is continuous for all $x > 4$ . $\newline$ Find $g(5)$ .

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Q. Let

g(x) = \frac{x-5}{\sqrt{x-4}-1}

when

x \neq 5

g

is continuous for all

x > 4

\newline

Find

g(5)

Recognize Indeterminate Form: To find the value of $g(5)$ , we need to evaluate the limit of $g(x)$ as $x$ approaches $5$ , because the function is not defined at $x=5$ but is continuous for $x > 4$ . We will use the limit process to find $g(5)$ .
Rationalize Denominator: First, we recognize that direct substitution of $x=5$ into $g(x)$ results in an indeterminate form $(0/0)$ . To resolve this, we can multiply the numerator and the denominator by the conjugate of the denominator to rationalize it.
Multiply by Conjugate: The conjugate of the denominator $\sqrt{x-4}-1$ is $\sqrt{x-4}+1$ . We multiply both the numerator and the denominator by this conjugate: $\newline$ $g(x) = \frac{(x-5)(\sqrt{x-4}+1)}{(\sqrt{x-4}-1)(\sqrt{x-4}+1)}$ .
Simplify Denominator: Simplifying the denominator using the difference of squares, we get: $g(x) = \frac{(x-5)(\sqrt{x-4}+1)}{(x-4) - 1}$ .
Cancel Terms: Further simplifying the denominator, we have: $\newline$ g(x) = $[\$x-5$ $\sqrt{x-4}+1$ ] / [x - $5$ ]\).
Final Simplification: Now, we can see that the $(x-5)$ terms in the numerator and denominator will cancel out, as long as $x$ is not equal to $5$ . This cancellation is valid because we are considering the limit as $x$ approaches $5$ , not the value at $x=5$ .
Substitute $x=5$ : After canceling out the $(x-5)$ terms, we are left with: $\newline$ g(x) = $\sqrt{x-4} + 1$ .
Calculate $g(5)$ : Now we can safely substitute $x=5$ into the simplified function to find the limit as $x$ approaches $5$ : $g(5) = \sqrt{5-4} + 1 = \sqrt{1} + 1 = 1 + 1$ .
Calculate $g(5)$ : Now we can safely substitute $x=5$ into the simplified function to find the limit as $x$ approaches $5$ : $g(5) = \sqrt{5-4} + 1 = \sqrt{1} + 1 = 1 + 1$ .Therefore, $g(5) = 2$ . This is the value of the function $g(x)$ as $x$ approaches $5$ .