Evaluate `lim_(x to 0) (tan(sgn(x)))/(sgn(x))` if exists.

Explanation

To evaluate the limit \( \lim_{x \to 0} \frac{\tan(\text{sgn}(x))}{\text{sgn}(x)} \), we will analyze the left-hand limit and the right-hand limit separately.

### Step 1: Understand the Signum Function
The signum function, denoted as \(\text{sgn}(x)\), is defined as:
- \(\text{sgn}(x) = 1\) if \(x > 0\)
- \(\text{sgn}(x) = 0\) if \(x = 0\)
- \(\text{sgn}(x) = -1\) if \(x < 0\)

### Step 2: Evaluate the Left-Hand Limit
We first evaluate the left-hand limit as \(x\) approaches \(0\) from the negative side:
\[ \lim_{x \to 0^-} \frac{\tan(\text{sgn}(x))}{\text{sgn}(x)} \]
For \(x < 0\), \(\text{sgn}(x) = -1\). Therefore, we have:
\[ \lim_{x \to 0^-} \frac{\tan(-1)}{-1} \]
This simplifies to:
\[ \frac{\tan(-1)}{-1} = -\tan(-1) = \tan(1) \]

### Step 3: Evaluate the Right-Hand Limit
Next, we evaluate the right-hand limit as \(x\) approaches \(0\) from the positive side:
\[ \lim_{x \to 0^+} \frac{\tan(\text{sgn}(x))}{\text{sgn}(x)} \]
For \(x > 0\), \(\text{sgn}(x) = 1\). Therefore, we have:
\[ \lim_{x \to 0^+} \frac{\tan(1)}{1} = \tan(1) \]

### Step 4: Compare the Left-Hand and Right-Hand Limits
Now we compare the two limits:
- Left-hand limit: \(\tan(1)\)
- Right-hand limit: \(\tan(1)\)

Since both limits are equal, we conclude that:
\[ \lim_{x \to 0} \frac{\tan(\text{sgn}(x))}{\text{sgn}(x)} = \tan(1) \]

### Final Result
Thus, the limit exists and is given by:
\[ \lim_{x \to 0} \frac{\tan(\text{sgn}(x))}{\text{sgn}(x)} = \tan(1) \]