The function `f:""R""~""{0}vec` given by `f(x)=1/x-2/(e^(2x)-1)` can be made continuous at x = 0 by defining f(0) as

Options

A. 0
B. 1
C. 2
D. `-1`

Explanation

To determine the value of \( f(0) \) that makes the function \( f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1} \) continuous at \( x = 0 \), we need to find the limit of \( f(x) \) as \( x \) approaches 0.

### Step 1: Set up the limit
We need to compute:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{1}{x} - \frac{2}{e^{2x} - 1} \right) \]

### Step 2: Rewrite the limit
To combine the two fractions, we find a common denominator:
\[ f(x) = \frac{(e^{2x} - 1) - 2x}{x(e^{2x} - 1)} \]
Thus, we rewrite the limit as:
\[ \lim_{x \to 0} \frac{(e^{2x} - 1) - 2x}{x(e^{2x} - 1)} \]

### Step 3: Evaluate the limit
As \( x \) approaches 0, both the numerator and denominator approach 0, which gives us a \( \frac{0}{0} \) indeterminate form. We can apply L'Hôpital's Rule.

First, we differentiate the numerator and the denominator:
- The derivative of the numerator \( (e^{2x} - 1) - 2x \) is \( 2e^{2x} - 2 \).
- The derivative of the denominator \( x(e^{2x} - 1) \) is \( (e^{2x} - 1) + 2xe^{2x} \).

Now we apply L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{2e^{2x} - 2}{(e^{2x} - 1) + 2xe^{2x}} \]

### Step 4: Substitute \( x = 0 \)
Now we substitute \( x = 0 \):
- The numerator becomes \( 2e^{0} - 2 = 2 - 2 = 0 \).
- The denominator becomes \( (e^{0} - 1) + 2 \cdot 0 \cdot e^{0} = 0 + 0 = 0 \).

Since we still have a \( \frac{0}{0} \) form, we apply L'Hôpital's Rule again.

### Step 5: Differentiate again
Differentiate the numerator and denominator again:
- The second derivative of the numerator \( 2e^{2x} \) is \( 4e^{2x} \).
- The second derivative of the denominator \( (e^{2x} - 1) + 2xe^{2x} \) is \( 2e^{2x} + 2e^{2x} + 4xe^{2x} = 4e^{2x} + 4xe^{2x} \).

Now we apply L'Hôpital's Rule again:
\[ \lim_{x \to 0} \frac{4e^{2x}}{4e^{2x} + 4xe^{2x}} \]

### Step 6: Substitute \( x = 0 \) again
Substituting \( x = 0 \):
- The numerator becomes \( 4e^{0} = 4 \).
- The denominator becomes \( 4e^{0} + 0 = 4 + 0 = 4 \).

Thus, the limit is:
\[ \lim_{x \to 0} \frac{4}{4} = 1 \]

### Conclusion
To make the function continuous at \( x = 0 \), we define:
\[ f(0) = 1 \]