Question

Let \[ f(x) = \frac{|5 - x|(x + 5)}{\tan(x - 5)} \quad {for} \quad x \neq 5. \] Then \[ \lim_{x \to 5} f(x) { is equal to:} \]

• A. 10
• B. -10
• C. 5
• D. -5
• E. 0

Hint:

For limits involving absolute values and trigonometric functions, rewrite the expression carefully and use standard limits such as \( \lim_{y \to 0} \frac{\tan y}{y} = 1 \) to simplify the evaluation process.

Answer 1

The Correct Option is A

We are asked to evaluate the following limit: \[ \lim_{x \to 5} f(x) = \lim_{x \to 5} \frac{|5 - x|(x + 5)}{\tan(x - 5)} \] Step 1: Simplifying the expression
First, we notice that \( |5 - x| \) depends on whether \( x \) is greater than or less than 5. As we are taking the limit as \( x \to 5 \), the value of \( |5 - x| \) will approach 0. So, we focus on the behavior near \( x = 5 \).
As \( x \) approaches 5, the expression \( (x - 5) \) in the denominator suggests that we are dealing with a limit involving \( \tan(x - 5) \). We recall the standard limit: \[ \lim_{y \to 0} \frac{\tan y}{y} = 1 \] Thus, we have: \[ \lim_{x \to 5} \frac{|5 - x|(x + 5)}{\tan(x - 5)} = \lim_{x \to 5} \frac{|5 - x|(x + 5)}{x - 5} \cdot \frac{x - 5}{\tan(x - 5)} = \lim_{x \to 5} |5 - x|(x + 5) \cdot \frac{1}{x - 5} \] Step 2: Applying the limit
Now, let's evaluate the limit:
- As \( x \to 5 \), \( |5 - x| \) becomes \( 0 \).
- The term \( (x + 5) \) approaches \( 10 \).
Thus, we have: \[ \lim_{x \to 5} |5 - x|(x + 5) = 0 \cdot 10 = 10 \] Thus, the correct answer is option (A), 10.