The jump value of the function at the point of the discontinuity of the function `f(x)=(1-k^(1//x))/(1+k^(1//x))(kgt0)` is

Options

A. 4
B. 2
C. 3
D. None of these

Answer

To find the jump value of the function \( f(x) = \frac{1 - k^{1/x}}{1 + k^{1/x}} \) at the point of discontinuity, we will calculate the left-hand limit and the right-hand limit as \( x \) approaches 0.

### Step 1: Define the limits
We need to find:
- Left-hand limit as \( x \to 0^- \)
- Right-hand limit as \( x \to 0^+ \)

### Step 2: Calculate the left-hand limit
For the left-hand limit, we evaluate:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1 - k^{1/x}}{1 + k^{1/x}} \]
As \( x \to 0^- \), \( \frac{1}{x} \to -\infty \) which implies \( k^{1/x} \to 0 \) (since \( k > 0 \)). Therefore, we have:
\[ \lim_{x \to 0^-} f(x) = \frac{1 - 0}{1 + 0} = 1 \]

### Step 3: Calculate the right-hand limit
For the right-hand limit, we evaluate:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{1 - k^{1/x}}{1 + k^{1/x}} \]
As \( x \to 0^+ \), \( \frac{1}{x} \to +\infty \) which implies \( k^{1/x} \to +\infty \). Thus, we can rewrite the limit as:
\[ \lim_{x \to 0^+} f(x) = \frac{1 - \infty}{1 + \infty} = \frac{-\infty}{\infty} \]
To simplify this, we can factor out \( k^{1/x} \):
\[ = \lim_{x \to 0^+} \frac{-k^{1/x}(1 - \frac{1}{k^{1/x}})}{k^{1/x}(1 + \frac{1}{k^{1/x}})} = \lim_{x \to 0^+} \frac{-1 + 0}{1 + 0} = -1 \]

### Step 4: Calculate the jump value
The jump value is given by:
\[ \text{Jump value} = \text{Left-hand limit} - \text{Right-hand limit} = 1 - (-1) = 1 + 1 = 2 \]

### Final Answer
Thus, the jump value of the function at the point of discontinuity is:
\[ \boxed{2} \]