Let $f(x) = \frac{\sqrt{x-1} - 2}{x-5}$ when $x \neq 5$ . $\newline$ $f$ is continuous for all $x > 1$ . $\newline$ Find $f(5)$ .

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Q. Let

f(x) = \frac{\sqrt{x-1} - 2}{x-5}

when

x \neq 5

\newline

f

is continuous for all

x > 1

\newline

Find

f(5)

Recognize Function Not Defined: First, we need to recognize that the function $f(x)$ is not defined at $x=5$ because it would result in a division by zero. However, we are asked to find $f(5)$ , which suggests we need to find the limit of $f(x)$ as $x$ approaches $5$ . This is because the function is continuous for all $x > 1$ except at $x=5$ .
Apply Limit Laws and Algebra: To find the limit of $f(x)$ as $x$ approaches $5$ , we can apply the limit laws and algebraic manipulation to simplify the expression and avoid the division by zero.
Simplify by Multiplying Conjugate: We will try to simplify the expression by multiplying the numerator and the denominator by the conjugate of the numerator. The conjugate of the numerator $\sqrt{x-1}-2$ is $\sqrt{x-1}+2$ .
Cancel Out Common Terms: Multiplying the numerator and denominator by the conjugate, we get: $\newline$ $\frac{(\sqrt{x-1}-2)(\sqrt{x-1}+2)}{(x-5)(\sqrt{x-1}+2)}$ $\newline$ This simplifies to: $\newline$ $\frac{(x-1) - 4}{(x-5)(\sqrt{x-1}+2)}$ $\newline$ Which further simplifies to: $\newline$ $\frac{x - 5}{(x-5)(\sqrt{x-1}+2)}$
Final Simplification: We can now cancel out the $(x-5)$ term in the numerator and the denominator, as long as $x \neq 5$ . This is valid because we are interested in the limit as $x$ approaches $5$ , not the value at $x=5$ .
Substitute $x$ with $5$ : After canceling out the $(x-5)$ term, we are left with: $\newline$ $\frac{1}{(\sqrt{x-1}+2)}$
Find Limit as x Approaches $5$ : Now we can substitute x with $5$ to find the limit of f(x) as x approaches $5$ : $\newline$ $\frac{1}{\sqrt{5-1}+2}$ $\newline$ $\frac{1}{\sqrt{4}+2}$ $\newline$ $\frac{1}{2+2}$ $\newline$ $\frac{1}{4}$
Conclusion: The limit of $f(x)$ as $x$ approaches $5$ is $\frac{1}{4}$ . Therefore, $f(5)$ is $\frac{1}{4}$ .