The function `f: R//{0} to R` is given by `f(x) = 1/x -2/(e^(2x)-1)` can be made continous at x=0 by defining f(0) as:

Options

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

Answer

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-by-Step Solution

1. **Define the function**:
\[ f(x) = \frac{1}{x} - \frac{2}{e^{2x} - 1} \]

2. **Find the limit as \( x \) approaches 0**:
We need to compute:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{1}{x} - \frac{2}{e^{2x} - 1} \right) \]

3. **Evaluate the limit of the second term**:
As \( x \to 0 \), \( e^{2x} \) can be approximated using the Taylor series expansion:
\[ e^{2x} \approx 1 + 2x + \frac{(2x)^2}{2} + \ldots = 1 + 2x + 2x^2 + \ldots \]
Therefore,
\[ e^{2x} - 1 \approx 2x + 2x^2 \]

4. **Substituting this back into the limit**:
\[ \lim_{x \to 0} \left( \frac{1}{x} - \frac{2}{2x + 2x^2} \right) \]
Simplifying the second term:
\[ \frac{2}{2x + 2x^2} = \frac{1}{x + x^2} = \frac{1}{x(1 + x)} \]

5. **Combine the terms**:
Now we have:
\[ \lim_{x \to 0} \left( \frac{1}{x} - \frac{1}{x(1 + x)} \right) \]

6. **Finding a common denominator**:
The common denominator is \( x(1 + x) \):
\[ \lim_{x \to 0} \left( \frac{(1 + x) - 1}{x(1 + x)} \right) = \lim_{x \to 0} \frac{x}{x(1 + x)} = \lim_{x \to 0} \frac{1}{1 + x} \]

7. **Evaluate the limit**:
As \( x \to 0 \):
\[ \lim_{x \to 0} \frac{1}{1 + x} = 1 \]

8. **Define \( f(0) \)**:
To make \( f(x) \) continuous at \( x = 0 \), we define:
\[ f(0) = 1 \]

### Conclusion
Thus, the function can be made continuous at \( x = 0 \) by defining \( f(0) = 1 \).