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 need to verify that:

\[ \lim_{x \to \pi} f(x) = f(\pi) \]

This means we will calculate the left-hand limit, the right-hand limit, and the value of the function at \( x = \pi \).

### Step 1: Calculate \( f(\pi) \)

First, we find the value of the function at \( x = \pi \):

\[ f(\pi) = |\sin(\pi) + \cos(\pi)| \]

Using the values of sine and cosine at \( \pi \):

\[ \sin(\pi) = 0 \quad \text{and} \quad \cos(\pi) = -1 \]

Thus,

\[ f(\pi) = |0 - 1| = | -1 | = 1 \]

### Step 2: Calculate the left-hand limit as \( x \) approaches \( \pi \)

Next, we calculate the left-hand limit:

\[ \lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} |\sin x + \cos x| \]

As \( x \) approaches \( \pi \) from the left, we can express \( x \) as \( \pi - h \) where \( h \to 0^+ \):

\[ \sin(\pi - h) = \sin h \quad \text{and} \quad \cos(\pi - h) = -\cos h \]

Thus,

\[ \sin x + \cos x = \sin(\pi - h) + \cos(\pi - h) = \sin h - \cos h \]

Now, taking the limit:

\[ \lim_{h \to 0} |\sin h - \cos h| \]

As \( h \to 0 \):

\[ \sin h \to 0 \quad \text{and} \quad \cos h \to 1 \]

So,

\[ \sin h - \cos h \to 0 - 1 = -1 \]

Thus,

\[ \lim_{h \to 0} |\sin h - \cos h| = |-1| = 1 \]

### Step 3: Calculate the right-hand limit as \( x \) approaches \( \pi \)

Now, we calculate the right-hand limit:

\[ \lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} |\sin x + \cos x| \]

As \( x \) approaches \( \pi \) from the right, we can express \( x \) as \( \pi + h \) where \( h \to 0^+ \):

\[ \sin(\pi + h) = -\sin h \quad \text{and} \quad \cos(\pi + h) = -\cos h \]

Thus,

\[ \sin x + \cos x = \sin(\pi + h) + \cos(\pi + h) = -\sin h - \cos h \]

Now, taking the limit:

\[ \lim_{h \to 0} |-\sin h - \cos h| \]

As \( h \to 0 \):

\[ -\sin h \to 0 \quad \text{and} \quad -\cos h \to -1 \]

So,

\[ -\sin h - \cos h \to 0 - 1 = -1 \]

Thus,

\[ \lim_{h \to 0} |-\sin h - \cos h| = |-1| = 1 \]

### Step 4: Conclusion

Now we have:

\[ \lim_{x \to \pi^-} f(x) = 1 \]
\[ \lim_{x \to \pi^+} f(x) = 1 \]
\[ f(\pi) = 1 \]

Since:

\[ \lim_{x \to \pi} f(x) = f(\pi) \]

We conclude that the function \( f(x) = |\sin x + \cos x| \) is continuous at \( x = \pi \).