`Lim_(x to1) (1-1/x)/(sin pi (x-1))`

Answer

To solve the limit problem \( \lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin(\pi (x - 1))} \), we will follow these steps:

### Step 1: Substitute the limit value
We start by substituting \( x = 1 \) into the expression:

\[ \frac{1 - \frac{1}{1}}{\sin(\pi (1 - 1))} = \frac{1 - 1}{\sin(0)} = \frac{0}{0} \]

This gives us an indeterminate form \( \frac{0}{0} \), which means we can apply L'Hôpital's Rule.

**Hint:** If substituting the limit gives \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), consider using L'Hôpital's Rule.

### Step 2: Differentiate the numerator and denominator
According to L'Hôpital's Rule, we differentiate the numerator and denominator separately.

- **Numerator:** \( 1 - \frac{1}{x} \)
- The derivative is \( 0 - \left(-\frac{1}{x^2}\right) = \frac{1}{x^2} \).

- **Denominator:** \( \sin(\pi (x - 1)) \)
- Using the chain rule, the derivative is \( \pi \cos(\pi (x - 1)) \).

### Step 3: Rewrite the limit using derivatives
Now we can rewrite the limit using the derivatives we found:

\[ \lim_{x \to 1} \frac{\frac{1}{x^2}}{\pi \cos(\pi (x - 1))} \]

### Step 4: Substitute the limit value again
Now we substitute \( x = 1 \) into the new expression:

\[ \frac{\frac{1}{1^2}}{\pi \cos(\pi (1 - 1))} = \frac{1}{\pi \cos(0)} = \frac{1}{\pi \cdot 1} = \frac{1}{\pi} \]

### Final Answer
Thus, the limit is:

\[ \lim_{x \to 1} \frac{1 - \frac{1}{x}}{\sin(\pi (x - 1))} = \frac{1}{\pi} \]

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