If `f(x) = sin (e^(x-2)-1)/(log (x-1))` then `lim_(x to 2)(f(x))` is given

Options

A. `-2`
B. `-1`
C. 0
D. 1

Answer

To find the limit of the function \( f(x) = \frac{\sin(e^{x-2} - 1)}{\log(x-1)} \) as \( x \) approaches 2, we can follow these steps:

### Step 1: Substitute the limit into the function
First, we substitute \( x = 2 \) into the function:
\[ f(2) = \frac{\sin(e^{2-2} - 1)}{\log(2-1)} = \frac{\sin(e^0 - 1)}{\log(1)} = \frac{\sin(0)}{0} \]
This results in the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
where \( f(x) \) is the numerator and \( g(x) \) is the denominator.

### Step 3: Differentiate the numerator and denominator
1. **Numerator**: \( f(x) = \sin(e^{x-2} - 1) \)
- The derivative is:
\[ f'(x) = \cos(e^{x-2} - 1) \cdot \frac{d}{dx}(e^{x-2} - 1) = \cos(e^{x-2} - 1) \cdot e^{x-2} \]

2. **Denominator**: \( g(x) = \log(x-1) \)
- The derivative is:
\[ g'(x) = \frac{1}{x-1} \]

### Step 4: Rewrite the limit using derivatives
Now we can rewrite the limit:
\[ \lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{\cos(e^{x-2} - 1) \cdot e^{x-2}}{\frac{1}{x-1}} = \lim_{x \to 2} \cos(e^{x-2} - 1) \cdot e^{x-2} \cdot (x-1) \]

### Step 5: Substitute \( x = 2 \) again
Now we substitute \( x = 2 \) into the new limit:
\[ = \cos(e^{2-2} - 1) \cdot e^{2-2} \cdot (2-1) = \cos(0) \cdot e^0 \cdot 1 = 1 \cdot 1 \cdot 1 = 1 \]

### Final Answer
Thus, we conclude that:
\[ \lim_{x \to 2} f(x) = 1 \]

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