Evaluate the following limits :
`Lim_(x to 0)(e^(x)-x-1)/x `

Answer

To evaluate the limit

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

we will follow these steps:

### Step 1: Rewrite the Limit
We can rewrite the limit as:

\[ \lim_{x \to 0} \left( \frac{e^x - 1}{x} - 1 \right). \]

This is because we can separate the expression into two parts:

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

### Step 2: Evaluate the First Part
Now, we need to evaluate the limit of the first part:

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

When we substitute \(x = 0\), we get:

\[ \frac{e^0 - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}, \]

which is an indeterminate form. Therefore, we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:

- The derivative of the numerator \(e^x - 1\) is \(e^x\).
- The derivative of the denominator \(x\) is \(1\).

Thus, we have:

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

### Step 4: Combine the Results
Now we can substitute back into our rewritten limit:

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

### Final Result
Therefore, the limit is:

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

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