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

Answer

To evaluate the limit

\[ \lim_{x \to 1} \frac{e^x - e}{x - 1}, \]

we can follow these steps:

### Step 1: Rewrite the limit expression

We can factor out \( e \) from the expression \( e^x - e \):

\[ e^x - e = e(e^{x-1} - 1). \]

Thus, we can rewrite the limit as:

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

### Step 2: Simplify the limit

Now, we can take \( e \) out of the limit since it is a constant:

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

### Step 3: Recognize the form of the limit

The limit

\[ \lim_{x \to 1} \frac{e^{x-1} - 1}{x - 1} \]

is of the indeterminate form \( \frac{0}{0} \) as both the numerator and the denominator approach 0 when \( x \) approaches 1.

### Step 4: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can use L'Hôpital's Rule, which states that:

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}, \]

if the limit on the right exists.

Here, let \( f(x) = e^{x-1} - 1 \) and \( g(x) = x - 1 \).

Calculating the derivatives:

- \( f'(x) = e^{x-1} \cdot 1 = e^{x-1} \)
- \( g'(x) = 1 \)

Now applying L'Hôpital's Rule:

\[ \lim_{x \to 1} \frac{e^{x-1}}{1} = e^{1-1} = e^0 = 1. \]

### Step 5: Substitute back into the limit

Now substituting back into our limit:

\[ = e \cdot 1 = e. \]

### Final Answer

Thus, the value of the limit is

\[ \lim_{x \to 1} \frac{e^x - e}{x - 1} = e. \]

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