Question

Test whether the function defined by \( f(x) = x^2 - \sin(x) + 5 \) is continuous at \( x = \pi \).

Hint:

Remember that polynomial functions (like \( x^2+5 \)) and basic trigonometric functions (like \( \sin x \)) are continuous everywhere in their domain. The sum, difference, and product of continuous functions are also continuous. This allows for a quick check without formal limit calculations in many cases.

Answer 1

Step 1: Understanding the Concept:
A function \( f(x) \) is said to be continuous at a point \( x = a \) if three conditions are met:
1. \( f(a) \) is defined.
2. The limit of the function as \( x \) approaches \( a \), \( \lim_{x \to a} f(x) \), exists.
3. The limit equals the function's value: \( \lim_{x \to a} f(x) = f(a) \).
Step 2: Detailed Explanation:
The given function is \( f(x) = x^2 - \sin(x) + 5 \). We need to test its continuity at \( x = \pi \).
Condition 1: Check if \( f(\pi) \) is defined.
Substitute \( x = \pi \) into the function:
\[ f(\pi) = (\pi)^2 - \sin(\pi) + 5 \] Since \( \sin(\pi) = 0 \), we get:
\[ f(\pi) = \pi^2 - 0 + 5 = \pi^2 + 5 \] The function is defined at \( x = \pi \).
Condition 2: Check if the limit exists.
We need to evaluate \( \lim_{x \to \pi} f(x) \).
\[ \lim_{x \to \pi} (x^2 - \sin(x) + 5) \] We can apply the limit to each term separately:
\[ \lim_{x \to \pi} x^2 - \lim_{x \to \pi} \sin(x) + \lim_{x \to \pi} 5 \] \[ = (\pi)^2 - \sin(\pi) + 5 = \pi^2 - 0 + 5 = \pi^2 + 5 \] The limit exists.
Condition 3: Compare the limit and function value.
From our calculations, we have:
\[ \lim_{x \to \pi} f(x) = \pi^2 + 5 \quad \text{and} \quad f(\pi) = \pi^2 + 5 \] Since \( \lim_{x \to \pi} f(x) = f(\pi) \), all three conditions for continuity are met.
Step 3: Final Answer:
The function \( f(x) = x^2 - \sin(x) + 5 \) is continuous at \( x = \pi \).