Evaluate the following `lim_(t to 1)(1-1/t)/(sin[pi(t-1)])`

Answer

To evaluate the limit

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

we can follow these steps:

### Step 1: Substitute \( t = 1 \)

First, we substitute \( t = 1 \) into the expression:

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

This results in an indeterminate form \( \frac{0}{0} \), so we need to simplify the expression.

### Step 2: Simplify the numerator

The numerator can be rewritten as:

\[ 1 - \frac{1}{t} = \frac{t - 1}{t} \]

Thus, we can rewrite the limit as:

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

### Step 3: Rewrite the limit

This gives us:

\[ \lim_{t \to 1} \frac{t - 1}{t \cdot \sin(\pi(t - 1))} \]

### Step 4: Substitute \( x = t - 1 \)

Let \( x = t - 1 \), then as \( t \to 1 \), \( x \to 0 \) and \( t = x + 1 \). The limit now becomes:

\[ \lim_{x \to 0} \frac{x}{(x + 1) \sin(\pi x)} \]

### Step 5: Evaluate the limit

Now, we can evaluate the limit:

\[ \lim_{x \to 0} \frac{x}{(x + 1) \sin(\pi x)} \]

Using the fact that \( \sin(\pi x) \approx \pi x \) as \( x \to 0 \), we can rewrite the limit:

\[ \lim_{x \to 0} \frac{x}{(x + 1) \cdot \pi x} = \lim_{x \to 0} \frac{1}{\pi(x + 1)} \]

### Step 6: Substitute \( x = 0 \)

Now substituting \( x = 0 \):

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

### Final Answer

Thus, the limit evaluates to:

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

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