Let `f(x)={{:((2^((1)/(x))-1)/(2^((1)/(x))+1),":",xne0),(0,":,x=0):}`, then `f(x)` is

Options

A. continuous and 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:

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

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

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

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

#### Left-Hand Limit (LHL)

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

As \( h \to 0^- \), \( -h \to 0^+ \) and \( \frac{1}{-h} \to -\infty \):

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

#### Right-Hand Limit (RHL)

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

As \( h \to 0^+ \), \( \frac{1}{h} \to +\infty \):

\[ = \lim_{h \to 0^+} \frac{\infty - 1}{\infty + 1} = \frac{\infty}{\infty} \text{ (indeterminate form)} \]

To resolve this, we can factor \( 2^{\frac{1}{h}} \) out of the numerator and denominator:

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

### Step 2: Compare Limits with \( f(0) \)

Now we compare the limits with \( f(0) \):

\[ f(0) = 0 \]

Since:

\[ \text{LHL} = -1 \quad \text{and} \quad \text{RHL} = 1 \quad \text{and} \quad f(0) = 0 \]

The left-hand limit and right-hand limit are not equal, thus:

\[ \text{LHL} \neq \text{RHL} \quad \Rightarrow \quad f \text{ is not continuous at } x = 0 \]

### Step 3: Conclusion on Differentiability

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

### Final Answer

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