If f(x)=`{{:(,(1)/(x)-(2)/(e^(2x)-1),x ne 0),(,1,x=0):}`

Options

A. f(x) is differentiable at x=0
B. f(x) is not differentiable at x=0
C. `f'(0)=(1)/(3)`
D. f(x) is continuous but not differenitable at x=0

Explanation

To determine the continuity and differentiability of the function

\[ f(x) = \begin{cases} \frac{1}{x} - \frac{2}{e^{2x} - 1} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \]

we will follow these steps:

### Step 1: Check Continuity at \( x = 0 \)

To check if \( f(x) \) is continuous at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0 and compare it to \( f(0) \).

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{1}{x} - \frac{2}{e^{2x} - 1} \right) \]

### Step 2: Simplify the Limit Expression

To simplify the limit, we can find a common denominator:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{(e^{2x} - 1) - 2x}{x(e^{2x} - 1)} \]

### Step 3: Expand \( e^{2x} \) Using Taylor Series

Using the Taylor series expansion for \( e^{2x} \):

\[ e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \ldots = 1 + 2x + 2x^2 + \frac{4x^3}{6} + \ldots \]

Thus,

\[ e^{2x} - 1 = 2x + 2x^2 + \frac{4x^3}{6} + \ldots \]

### Step 4: Substitute Back into the Limit

Now substituting back into our limit expression:

\[ \lim_{x \to 0} \frac{(2x + 2x^2 + \ldots) - 2x}{x(2x + 2x^2 + \ldots)} = \lim_{x \to 0} \frac{2x^2 + \ldots}{x(2x + 2x^2 + \ldots)} \]

### Step 5: Cancel Terms and Evaluate the Limit

Cancel \( x \) from numerator and denominator:

\[ = \lim_{x \to 0} \frac{2x + \ldots}{2 + 2x + \ldots} \]

As \( x \to 0 \), this simplifies to:

\[ = \frac{0}{2} = 0 \]

### Step 6: Compare with \( f(0) \)

Since \( f(0) = 1 \) and \( \lim_{x \to 0} f(x) = 0 \), we conclude that:

\[ \lim_{x \to 0} f(x) \neq f(0) \]

Thus, \( f(x) \) is **not continuous** at \( x = 0 \).

### Step 7: Check Differentiability at \( x = 0 \)

Since \( f(x) \) is not continuous at \( x = 0 \), it cannot be differentiable at that point. However, for completeness, we can check the left-hand derivative (LHD) and right-hand derivative (RHD).

**Left-Hand Derivative (LHD):**

\[ \text{LHD} = \lim_{h \to 0^-} \frac{f(0) - f(-h)}{-h} = \lim_{h \to 0^-} \frac{1 - \left( \frac{1}{-h} - \frac{2}{e^{-2h} - 1} \right)}{-h} \]

**Right-Hand Derivative (RHD):**

\[ \text{RHD} = \lim_{h \to 0^+} \frac{f(0) - f(h)}{-h} = \lim_{h \to 0^+} \frac{1 - \left( \frac{1}{h} - \frac{2}{e^{2h} - 1} \right)}{-h} \]

Both derivatives will yield different results, confirming that \( f(x) \) is not differentiable at \( x = 0 \).

### Conclusion

The function \( f(x) \) is neither continuous nor differentiable at \( x = 0 \).

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