Evaluate the following limits :
`Lim_(xto0) (e^(x)+e^(-x)-2)/x`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{x}, \]

we can follow these steps:

### Step 1: Substitute \( x = 0 \)

First, we substitute \( x = 0 \) into the expression:

\[ e^0 + e^{-0} - 2 = 1 + 1 - 2 = 0. \]

Thus, the limit becomes:

\[ \frac{0}{0}, \]

which is an indeterminate form.

**Hint:** When you encounter a \( \frac{0}{0} \) form, consider using L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if the limit results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can differentiate the numerator and the denominator separately.

Differentiate the numerator \( e^x + e^{-x} - 2 \):

- The derivative of \( e^x \) is \( e^x \).
- The derivative of \( e^{-x} \) is \( -e^{-x} \).
- The derivative of the constant \( -2 \) is \( 0 \).

So, the derivative of the numerator is:

\[ e^x - e^{-x}. \]

The derivative of the denominator \( x \) is \( 1 \).

Thus, we can rewrite the limit as:

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

**Hint:** After applying L'Hôpital's Rule, simplify the limit expression.

### Step 3: Substitute \( x = 0 \) Again

Now we substitute \( x = 0 \) into the new limit:

\[ e^0 - e^{-0} = 1 - 1 = 0. \]

Again, we have \( \frac{0}{1} \), which is not indeterminate.

**Hint:** If you still get \( 0 \), check if you can apply L'Hôpital's Rule again or simplify further.

### Step 4: Evaluate the Limit

Since we have \( e^x - e^{-x} \) as \( x \to 0 \):

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

Thus, the final limit is:

\[ \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{x} = 0. \]

### Final Answer

The limit evaluates to:

\[ \boxed{0}. \]