`lim_(xtooo) (e^(1//x^(2))-1)/(2tan^(-1)(x^(2))-pi)` is equal to

Options

A. `e^((1-e))`
B. `e^(((1-e)/e))`
C. `e^(((e)/(1-e)))`
D. `e^((1+e)/(e))`

Explanation

To solve the limit

\[ \lim_{x \to 0} \frac{e^{\frac{1}{x^2}} - 1}{2\tan^{-1}(x^2) - \pi}, \]

we will follow these steps:

### Step 1: Substitute \( x \) with \( \frac{1}{t} \)

Let \( t = \frac{1}{x} \), which means as \( x \to 0 \), \( t \to \infty \). Thus, we rewrite the limit:

\[ \lim_{t \to \infty} \frac{e^{t^2} - 1}{2\tan^{-1}\left(\frac{1}{t^2}\right) - \pi}. \]

### Step 2: Analyze the numerator and denominator

As \( t \to \infty \):

- The numerator \( e^{t^2} - 1 \) approaches \( e^{t^2} \).
- The denominator \( 2\tan^{-1}\left(\frac{1}{t^2}\right) - \pi \) approaches \( 2 \cdot 0 - \pi = -\pi \).

Thus, we can rewrite the limit as:

\[ \lim_{t \to \infty} \frac{e^{t^2}}{-\pi}. \]

### Step 3: Evaluate the limit

Since \( e^{t^2} \) grows exponentially as \( t \to \infty \), the limit becomes:

\[ \lim_{t \to \infty} \frac{e^{t^2}}{-\pi} = -\infty. \]

### Conclusion

The limit diverges to negative infinity:

\[ \lim_{x \to 0} \frac{e^{\frac{1}{x^2}} - 1}{2\tan^{-1}(x^2) - \pi} = -\infty. \]

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