`lim_(x rarr -pi) |x+pi|/sinx` is

Options

A. `-1`
B. `1`
C. `pi`
D. Does not exist

Answer

To solve the limit \( \lim_{x \to -\pi} \frac{|x + \pi|}{\sin x} \), we need to analyze the expression step by step.

### Step 1: Determine the behavior of \( |x + \pi| \) as \( x \to -\pi \)

As \( x \) approaches \(-\pi\), we can evaluate the expression \( |x + \pi| \):
- For \( x < -\pi \), \( x + \pi < 0 \) and thus \( |x + \pi| = -(x + \pi) \).
- For \( x > -\pi \), \( x + \pi > 0 \) and thus \( |x + \pi| = x + \pi \).

Since we are interested in the limit as \( x \) approaches \(-\pi\), we will consider both the left-hand limit and the right-hand limit.

### Step 2: Calculate the left-hand limit \( \lim_{x \to -\pi^-} \frac{|x + \pi|}{\sin x} \)

For \( x \to -\pi^- \):
\[ |x + \pi| = -(x + \pi) \]
So the limit becomes:
\[ \lim_{x \to -\pi^-} \frac{-(x + \pi)}{\sin x} \]

Substituting \( x = -\pi + h \) where \( h \to 0^- \):
\[ \lim_{h \to 0^-} \frac{-(-\pi + h + \pi)}{\sin(-\pi + h)} = \lim_{h \to 0^-} \frac{-h}{\sin(-\pi + h)} \]
Using the identity \( \sin(-\pi + h) = -\sin(h) \):
\[ = \lim_{h \to 0^-} \frac{-h}{-\sin(h)} = \lim_{h \to 0^-} \frac{h}{\sin(h)} \]

As \( h \to 0 \), we know that \( \frac{h}{\sin(h)} \to 1 \):
\[ \lim_{x \to -\pi^-} \frac{|x + \pi|}{\sin x} = 1 \]

### Step 3: Calculate the right-hand limit \( \lim_{x \to -\pi^+} \frac{|x + \pi|}{\sin x} \)

For \( x \to -\pi^+ \):
\[ |x + \pi| = x + \pi \]
So the limit becomes:
\[ \lim_{x \to -\pi^+} \frac{x + \pi}{\sin x} \]

Substituting \( x = -\pi + h \) where \( h \to 0^+ \):
\[ \lim_{h \to 0^+} \frac{(-\pi + h) + \pi}{\sin(-\pi + h)} = \lim_{h \to 0^+} \frac{h}{\sin(-\pi + h)} \]
Using the identity \( \sin(-\pi + h) = -\sin(h) \):
\[ = \lim_{h \to 0^+} \frac{h}{-\sin(h)} = \lim_{h \to 0^+} \frac{-h}{\sin(h)} \]

As \( h \to 0 \), we know that \( \frac{-h}{\sin(h)} \to -1 \):
\[ \lim_{x \to -\pi^+} \frac{|x + \pi|}{\sin x} = -1 \]

### Step 4: Conclusion

Since the left-hand limit is \( 1 \) and the right-hand limit is \( -1 \), we conclude:
\[ \lim_{x \to -\pi} \frac{|x + \pi|}{\sin x} \text{ does not exist.} \]