Let `y=-(2^(1/x)-1)/(2^(1/x)+1)` , then

Options

A. `lim_(xto0) y=-1`
B. `lim_(x to 0) y=1`
C. `lim_(xto0+) y=-1`
D. `lim_(x to 0+) y =lim_(x to 0-) y`

Answer

To solve the problem, we need to find the limits of the function \( y = -\frac{2^{1/x} - 1}{2^{1/x} + 1} \) as \( x \) approaches 0 from both the left and the right.

### Step-by-Step Solution:

1. **Identify the Function**:
We have \( y = -\frac{2^{1/x} - 1}{2^{1/x} + 1} \).

2. **Evaluate the Left-Hand Limit**:
We will first find the limit as \( x \) approaches 0 from the left (denoted as \( 0^- \)).
\[ \lim_{x \to 0^-} y = \lim_{h \to 0} -\frac{2^{1/(0-h)} - 1}{2^{1/(0-h)} + 1} = \lim_{h \to 0} -\frac{2^{-1/h} - 1}{2^{-1/h} + 1} \]

3. **Analyze \( 2^{-1/h} \)**:
As \( h \to 0^- \), \( -1/h \) approaches \( -\infty \), hence \( 2^{-1/h} \) approaches 0.
\[ \lim_{h \to 0} -\frac{0 - 1}{0 + 1} = -\frac{-1}{1} = 1 \]
Therefore,
\[ \lim_{x \to 0^-} y = 1 \]

4. **Evaluate the Right-Hand Limit**:
Next, we find the limit as \( x \) approaches 0 from the right (denoted as \( 0^+ \)).
\[ \lim_{x \to 0^+} y = \lim_{h \to 0} -\frac{2^{1/(0+h)} - 1}{2^{1/(0+h)} + 1} = \lim_{h \to 0} -\frac{2^{1/h} - 1}{2^{1/h} + 1} \]

5. **Analyze \( 2^{1/h} \)**:
As \( h \to 0^+ \), \( 1/h \) approaches \( +\infty \), hence \( 2^{1/h} \) approaches \( +\infty \).
\[ \lim_{h \to 0} -\frac{+\infty - 1}{+\infty + 1} = -\frac{\infty}{\infty} = -1 \]
Therefore,
\[ \lim_{x \to 0^+} y = -1 \]

6. **Conclusion**:
Since the left-hand limit \( \lim_{x \to 0^-} y = 1 \) and the right-hand limit \( \lim_{x \to 0^+} y = -1 \) are not equal, we conclude that the limit does not exist as \( x \) approaches 0.

### Final Result:
The limit \( \lim_{x \to 0} y \) does not exist.