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

Answer

To evaluate the limit

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

we will follow these steps:

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

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

\[ \frac{e^{1} - e^{-1}}{e^{1} + e^{-1}}. \]

### Step 2: Calculate \( e^{1} \) and \( e^{-1} \)

Now, we calculate \( e^{1} \) and \( e^{-1} \):

- \( e^{1} = e \)
- \( e^{-1} = \frac{1}{e} \)

### Step 3: Substitute these values into the expression

Substituting these values into the limit gives:

\[ \frac{e - \frac{1}{e}}{e + \frac{1}{e}}. \]

### Step 4: Simplify the numerator and denominator

Now, we simplify both the numerator and the denominator:

**Numerator:**

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

**Denominator:**

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

### Step 5: Substitute back into the limit

Now we can substitute these simplified forms back into the limit:

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

### Step 6: Cancel the common terms

The \( e \) in the numerator and denominator cancels out:

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

### Final Result

Thus, the limit evaluates to:

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

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