1.4 Q-6

Question Statement

Determine the value of $K$ such that the function

$$f(x) = \begin{cases} \frac{\sqrt{2x+5} - \sqrt{x+7}}{x-2}, & x \neq 2 K, & x = 2 \end{cases}$$

is continuous at $x = 2$.

Background and Explanation

To ensure $f(x)$ is continuous at $x = 2$, the following conditions must hold:

1. $f(2)$ must exist.
2. The left-hand limit ($\lim_{x \to 2^-} f(x)$) and the right-hand limit ($\lim_{x \to 2^+} f(x)$) must exist and be equal.
3. These limits must equal $f(2)$.

The challenge in this problem is to evaluate the given piecewise function’s limit as $x \to 2$, where direct substitution initially results in an indeterminate form $\frac{0}{0}$. We resolve this using algebraic manipulation (rationalizing the numerator).

Solution

Step 1: Define $f(2)$

The value of $f(2)$ is given as $K$.

$$f(2) = K$$

Step 2: Evaluate the Limit of $f(x)$ as $x \to 2$

For $x \neq 2$,

$$f(x) = \frac{\sqrt{2x+5} - \sqrt{x+7}}{x-2}$$

Direct substitution gives $\frac{0}{0}$, so we rationalize the numerator by multiplying and dividing by $\sqrt{2x+5} + \sqrt{x+7}$.

Rationalizing the Numerator:

$$\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{(\sqrt{2x+5})^2 - (\sqrt{x+7})^2}{(x-2)(\sqrt{2x+5} + \sqrt{x+7})}$$

Simplify the numerator using the difference of squares:

$$(\sqrt{2x+5})^2 - (\sqrt{x+7})^2 = (2x + 5) - (x + 7) = x - 2$$

The $x-2$ in the numerator cancels with the $x-2$ in the denominator:

$$\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{1}{\sqrt{2x+5} + \sqrt{x+7}}$$

Substituting $x = 2$ into the Remaining Expression:

$$\lim_{x \to 2} f(x) = \frac{1}{\sqrt{2(2) + 5} + \sqrt{2 + 7}} = \frac{1}{\sqrt{9} + \sqrt{9}}$$ $$\lim_{x \to 2} f(x) = \frac{1}{3 + 3} = \frac{1}{6}$$

Step 3: Apply the Continuity Condition

For continuity at $x = 2$, we must have:

$$f(2) = \lim_{x \to 2} f(x)$$

Substitute $\lim_{x \to 2} f(x) = \frac{1}{6}$ and $f(2) = K$:

$$K = \frac{1}{6}$$

Final Answer:

$$K = \frac{1}{6}$$

Key Formulas or Methods Used

1. Continuity Condition:
   $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$.

2. Rationalization Technique:
   Multiply and divide by $\sqrt{2x+5} + \sqrt{x+7}$ to simplify indeterminate forms.

3. Difference of Squares Formula:
   $a^2 - b^2 = (a-b)(a+b)$.

Summary of Steps

1. Define $f(2) = K$.
2. Evaluate $\lim_{x \to 2} f(x)$ by rationalizing the numerator.
3. Simplify using the difference of squares and cancel terms.
4. Substitute $x = 2$ into the simplified expression to find $\lim_{x \to 2} f(x) = \frac{1}{6}$.
5. Apply the continuity condition $K = \lim_{x \to 2} f(x)$ to find $K = \frac{1}{6}$.