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

Options

A. Continuous as well as differentiable at x = 0
B. Continuous but not differentiable at x = 0
C. Differentiable but not continuous at x = 0
D. None of these

Answer

To determine the nature of the function \( f(x) \) defined as:

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

we need to check if \( f(x) \) is continuous and differentiable at \( x = 0 \).

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

To check the continuity of \( f(x) \) at \( x = 0 \), we need to find the left-hand limit (LHL) and the right-hand limit (RHL) as \( x \) approaches 0.

#### Left-Hand Limit (LHL)

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

As \( x \) approaches 0 from the left (negative side), \( \frac{2}{x} \) approaches \( -\infty \). Therefore:

\[ e^{\frac{2}{x}} \to e^{-\infty} = 0 \]

Substituting this into the limit:

\[ \text{LHL} = \frac{0 - 1}{0 + 1} = \frac{-1}{1} = -1 \]

#### Right-Hand Limit (RHL)

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

As \( x \) approaches 0 from the right (positive side), \( \frac{2}{x} \) approaches \( +\infty \). Therefore:

\[ e^{\frac{2}{x}} \to e^{+\infty} = +\infty \]

Substituting this into the limit:

\[ \text{RHL} = \frac{+\infty - 1}{+\infty + 1} = \frac{+\infty}{+\infty} = 1 \]

### Step 2: Compare LHL and RHL

Since:

\[ \text{LHL} = -1 \quad \text{and} \quad \text{RHL} = 1 \]

The left-hand limit does not equal the right-hand limit:

\[ \text{LHL} \neq \text{RHL} \]

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

### Step 3: Check Differentiability at \( x = 0 \)

A function must be continuous at a point to be differentiable at that point. Since \( f(x) \) is not continuous at \( x = 0 \), it cannot be differentiable at \( x = 0 \).

### Conclusion

Since \( f(x) \) is neither continuous nor differentiable at \( x = 0 \), we conclude that:

**The function \( f(x) \) is discontinuous and non-differentiable at \( x = 0 \).**