Find the limit of the function f(x) if , it exists , defined as follows : `f(x) = (e^(1//x)-1)/(e^(1//x)+1) , x != 0 ` as x tends to zero .

Answer

To find the limit of the function \( f(x) = \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1} \) as \( x \) tends to 0, we can follow these steps:

### Step 1: Analyze the limit
We need to evaluate:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1} \]
As \( x \) approaches 0, \( \frac{1}{x} \) approaches \( \infty \) (from the right side, since \( x \) is positive). Therefore, \( e^{\frac{1}{x}} \) approaches \( \infty \).

### Step 2: Substitute the limit
Substituting \( e^{\frac{1}{x}} \) into the limit gives:
\[ \lim_{x \to 0} \frac{\infty - 1}{\infty + 1} = \frac{\infty}{\infty} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
L'Hôpital's Rule states that if we have an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can differentiate the numerator and the denominator separately.

Differentiate the numerator:
\[ \text{Numerator: } \frac{d}{dx}(e^{\frac{1}{x}} - 1) = e^{\frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right) = -\frac{e^{\frac{1}{x}}}{x^2} \]

Differentiate the denominator:
\[ \text{Denominator: } \frac{d}{dx}(e^{\frac{1}{x}} + 1) = e^{\frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right) = -\frac{e^{\frac{1}{x}}}{x^2} \]

### Step 4: Rewrite the limit using derivatives
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{-\frac{e^{\frac{1}{x}}}{x^2}}{-\frac{e^{\frac{1}{x}}}{x^2}} = \lim_{x \to 0} \frac{e^{\frac{1}{x}}}{e^{\frac{1}{x}}} = 1 \]

### Conclusion
Thus, the limit exists and is equal to:
\[ \lim_{x \to 0} f(x) = 1 \]