Applying the L.Hospital rule , find the limits of the following functions :
`lim_(x tooo) (e^(1//x^(2))-1)/(2 " arc tan" x^(2)-pi)`

Answer

To find the limit of the function

\[ \lim_{x \to \infty} \frac{e^{\frac{1}{x^2}} - 1}{2 \arctan(x^2) - \pi}, \]

we will apply L'Hospital's Rule since the limit results in the indeterminate form \( \frac{0}{0} \).

### Step-by-Step Solution:

1. **Check the form of the limit**:
- As \( x \to \infty \), \( e^{\frac{1}{x^2}} \) approaches \( e^0 = 1 \), thus \( e^{\frac{1}{x^2}} - 1 \) approaches \( 0 \).
- For \( 2 \arctan(x^2) - \pi \), as \( x \to \infty \), \( \arctan(x^2) \) approaches \( \frac{\pi}{2} \), so \( 2 \arctan(x^2) - \pi \) approaches \( 0 \).
- Therefore, we have the form \( \frac{0}{0} \).

2. **Apply L'Hospital's Rule**:
- Differentiate the numerator:
\[ \frac{d}{dx}(e^{\frac{1}{x^2}} - 1) = e^{\frac{1}{x^2}} \cdot \frac{d}{dx}\left(\frac{1}{x^2}\right) = e^{\frac{1}{x^2}} \cdot \left(-\frac{2}{x^3}\right) = -\frac{2 e^{\frac{1}{x^2}}}{x^3}. \]
- Differentiate the denominator:
\[ \frac{d}{dx}(2 \arctan(x^2) - \pi) = 2 \cdot \frac{1}{1 + (x^2)^2} \cdot \frac{d}{dx}(x^2) = 2 \cdot \frac{1}{1 + x^4} \cdot 2x = \frac{4x}{1 + x^4}. \]

3. **Rewrite the limit using derivatives**:
\[ \lim_{x \to \infty} \frac{-\frac{2 e^{\frac{1}{x^2}}}{x^3}}{\frac{4x}{1 + x^4}} = \lim_{x \to \infty} \frac{-2 e^{\frac{1}{x^2}} (1 + x^4)}{4x^4}. \]

4. **Simplify the expression**:
\[ = \lim_{x \to \infty} -\frac{1}{2} \cdot \frac{e^{\frac{1}{x^2}} (1 + x^4)}{x^4}. \]

5. **Evaluate the limit as \( x \to \infty \)**:
- As \( x \to \infty \), \( e^{\frac{1}{x^2}} \to 1 \) and \( \frac{1 + x^4}{x^4} \to 1 \).
- Thus, the limit becomes:
\[ = -\frac{1}{2} \cdot 1 \cdot 1 = -\frac{1}{2}. \]

### Final Answer:
\[ \lim_{x \to \infty} \frac{e^{\frac{1}{x^2}} - 1}{2 \arctan(x^2) - \pi} = -\frac{1}{2}. \]