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

Answer

To evaluate the limit \( \lim_{x \to 1} \frac{\sin(e^x - 1)}{\log x} \), we will follow these steps:

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

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

\[ \sin(e^1 - 1) = \sin(e - 1) \]
\[ \log(1) = 0 \]

This gives us the form \( \frac{\sin(e - 1)}{0} \), which is not defined. However, if we check the limit as \( x \) approaches 1, we see that both the numerator and denominator approach 0, leading to 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 we can take the derivative of the numerator and the denominator separately.

The limit now becomes:

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

### Step 3: Differentiate the Numerator and Denominator

**Numerator:**
Using the chain rule, the derivative of \( \sin(e^x - 1) \) is:

\[ \cos(e^x - 1) \cdot \frac{d}{dx}(e^x) = \cos(e^x - 1) \cdot e^x \]

**Denominator:**
The derivative of \( \log x \) is:

\[ \frac{1}{x} \]

### Step 4: Rewrite the Limit

Now we can rewrite the limit as:

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

### Step 5: Substitute \( x = 1 \) Again

Now we substitute \( x = 1 \) into the new expression:

\[ \cos(e^1 - 1) \cdot e^1 \cdot 1 = \cos(e - 1) \cdot e \]

### Final Result

Thus, the limit evaluates to:

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