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 properties 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 will check the continuity and differentiability of \( f(x) \) at \( x = 0 \).

### 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 with \( f(0) \).

#### Step 1.1: Calculate the Right-Hand Limit

The right-hand limit as \( x \) approaches 0 from the positive side is given by:

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

Let \( h = \frac{1}{x} \) so that as \( x \to 0^+ \), \( h \to +\infty \). Thus, we can rewrite the limit as:

\[ \lim_{h \to +\infty} \frac{e^{2h} - 1}{e^{2h} + 1} \]

As \( h \to +\infty \), \( e^{2h} \) dominates both the numerator and the denominator:

\[ \lim_{h \to +\infty} \frac{e^{2h} - 1}{e^{2h} + 1} = \lim_{h \to +\infty} \frac{1 - \frac{1}{e^{2h}}}{1 + \frac{1}{e^{2h}}} = \frac{1 - 0}{1 + 0} = 1 \]

#### Step 1.2: Calculate the Left-Hand Limit

Similarly, the left-hand limit as \( x \) approaches 0 from the negative side is:

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

Using the same substitution \( h = \frac{1}{x} \), as \( x \to 0^- \), \( h \to -\infty \):

\[ \lim_{h \to -\infty} \frac{e^{2h} - 1}{e^{2h} + 1} \]

As \( h \to -\infty \), \( e^{2h} \) approaches 0:

\[ \lim_{h \to -\infty} \frac{0 - 1}{0 + 1} = -1 \]

#### Step 1.3: Compare Limits with \( f(0) \)

Now we have:

- Right-hand limit: \( 1 \)
- Left-hand limit: \( -1 \)
- \( f(0) = 0 \)

Since the right-hand limit and left-hand limit are not equal, and neither of them equals \( f(0) \), we conclude that:

\[ f(x) \text{ is not continuous at } x = 0. \]

### Step 2: Check Differentiability at \( x = 0 \)

A function must be continuous at a point to be differentiable there. Since we have established that \( f(x) \) is not continuous at \( x = 0 \), it follows that:

\[ f(x) \text{ is not differentiable at } x = 0. \]

### Conclusion

Thus, the function \( f(x) \) is neither continuous nor differentiable at \( x = 0 \).