Show that the function ` f(x) = |sinx+cosx|` is continuous at `x = pi`.

Explanation

To show that the function \( f(x) = |\sin x + \cos x| \) is continuous at \( x = \pi \), we will follow these steps:

### Step 1: Understand the Function
We start with the function:
\[ f(x) = |\sin x + \cos x| \]
We need to check the continuity of this function at \( x = \pi \).

### Step 2: Check Continuity Definition
A function \( f(x) \) is continuous at a point \( c \) if:
\[ \lim_{x \to c} f(x) = f(c) \]
In our case, \( c = \pi \).

### Step 3: Calculate \( f(\pi) \)
First, we calculate \( f(\pi) \):
\[ f(\pi) = |\sin(\pi) + \cos(\pi)| = |0 - 1| = 1 \]

### Step 4: Calculate the Limit as \( x \) Approaches \( \pi \)
Next, we need to find the limit of \( f(x) \) as \( x \) approaches \( \pi \):
\[ \lim_{x \to \pi} f(x) = \lim_{x \to \pi} |\sin x + \cos x| \]

### Step 5: Evaluate \( \sin x + \cos x \) Near \( x = \pi \)
We can evaluate \( \sin x + \cos x \) as \( x \) approaches \( \pi \):
\[ \sin x \text{ approaches } 0 \text{ and } \cos x \text{ approaches } -1 \]
Thus,
\[ \sin x + \cos x \to 0 - 1 = -1 \]

### Step 6: Take the Absolute Value
Now, we take the absolute value:
\[ \lim_{x \to \pi} |\sin x + \cos x| = |-1| = 1 \]

### Step 7: Compare Limit with Function Value
Now we compare the limit with the function value:
\[ \lim_{x \to \pi} f(x) = 1 \quad \text{and} \quad f(\pi) = 1 \]

### Step 8: Conclusion
Since:
\[ \lim_{x \to \pi} f(x) = f(\pi) \]
We conclude that \( f(x) \) is continuous at \( x = \pi \).

### Final Result
Thus, we have shown that the function \( f(x) = |\sin x + \cos x| \) is continuous at \( x = \pi \).

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