Evaluate `lim_(xto0) (e^(x)+e^(-x)-2)/(x^(2))`

Explanation

To evaluate the limit

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

we first notice that substituting \( x = 0 \) directly into the expression gives us the form \( \frac{0}{0} \), which is indeterminate. Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and the derivative of the denominator separately.

### Step 1: Differentiate the numerator and denominator

The numerator is \( e^x + e^{-x} - 2 \). The derivative of this is:

\[ \frac{d}{dx}(e^x + e^{-x} - 2) = e^x - e^{-x}. \]

The denominator is \( x^2 \). The derivative of this is:

\[ \frac{d}{dx}(x^2) = 2x. \]

Now we can rewrite our limit using these derivatives:

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

### Step 2: Evaluate the limit again

Substituting \( x = 0 \) into this new expression also gives us \( \frac{0}{0} \). We apply L'Hôpital's Rule again.

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

\[ \frac{d}{dx}(e^x - e^{-x}) = e^x + e^{-x}. \]

Differentiate the denominator \( 2x \):

\[ \frac{d}{dx}(2x) = 2. \]

Now we have:

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

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

Now we can substitute \( x = 0 \):

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

### Final Answer

Thus, the limit is:

\[ \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{x^2} = 1. \]