Evaluate the following limits :
`Lim_(xto2) (sin(e^(x-2)-1))/(log(x-1))`

Answer

To evaluate the limit

\[ \lim_{x \to 2} \frac{\sin(e^{x-2} - 1)}{\log(x - 1)}, \]

we will follow these steps:

### Step 1: Substitute the limit value
First, we substitute \( x = 2 \) into the limit expression:

\[ \sin(e^{2-2} - 1) = \sin(e^0 - 1) = \sin(1 - 1) = \sin(0) = 0, \]
and

\[ \log(2 - 1) = \log(1) = 0. \]

Thus, we have the form \( \frac{0}{0} \), which is indeterminate.

**Hint:** Always check the limit by direct substitution first to see if it results in an indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can differentiate the numerator and denominator separately.

We differentiate the numerator and denominator:

- The numerator:
\[ \frac{d}{dx}[\sin(e^{x-2} - 1)] = \cos(e^{x-2} - 1) \cdot \frac{d}{dx}[e^{x-2}] = \cos(e^{x-2} - 1) \cdot e^{x-2} \cdot 1 = e^{x-2} \cos(e^{x-2} - 1). \]

- The denominator:
\[ \frac{d}{dx}[\log(x - 1)] = \frac{1}{x - 1}. \]

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

\[ \lim_{x \to 2} \frac{e^{x-2} \cos(e^{x-2} - 1)}{\frac{1}{x - 1}} = \lim_{x \to 2} e^{x-2} \cos(e^{x-2} - 1) \cdot (x - 1). \]

### Step 4: Substitute the limit value again
Now we substitute \( x = 2 \):

- For \( e^{x-2} \):
\[ e^{2-2} = e^0 = 1. \]

- For \( \cos(e^{x-2} - 1) \):
\[ \cos(e^{2-2} - 1) = \cos(1 - 1) = \cos(0) = 1. \]

- For \( (x - 1) \):
\[ 2 - 1 = 1. \]

Putting it all together:

\[ \lim_{x \to 2} e^{x-2} \cos(e^{x-2} - 1) \cdot (x - 1) = 1 \cdot 1 \cdot 1 = 1. \]

### Final Answer
Thus, the limit evaluates to:

\[ \boxed{1}. \]