`Lt_(x to 0)(e^(1//x)-1)/(e^(1//x)+1)` is equal to:

Options

A. 1
B. `-1`
C. 0
D. does not exist

Answer

To solve the limit \( \lim_{x \to 0} \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1} \), we will follow these steps:

### Step 1: Analyze the behavior of \( e^{\frac{1}{x}} \) as \( x \to 0 \)
As \( x \) approaches \( 0 \) from the positive side, \( \frac{1}{x} \) approaches \( +\infty \). Therefore, \( e^{\frac{1}{x}} \) approaches \( +\infty \).

### Step 2: Substitute the limit into the expression
Now we can substitute this behavior into our limit:
\[ \lim_{x \to 0} \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1} = \lim_{x \to 0} \frac{+\infty - 1}{+\infty + 1} \]

### Step 3: Simplify the expression
As \( e^{\frac{1}{x}} \) becomes very large, the \( -1 \) and \( +1 \) become negligible:
\[ \lim_{x \to 0} \frac{+\infty - 1}{+\infty + 1} = \lim_{x \to 0} \frac{+\infty}{+\infty} \]

### Step 4: Apply the limit
This simplifies to:
\[ \lim_{x \to 0} \frac{e^{\frac{1}{x}}}{e^{\frac{1}{x}}} = 1 \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to 0} \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1} = 1 \]

### Final Answer
The limit is equal to \( 1 \).

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