If `L=lim_(xrarr0)((e^(-x^(2)/2)-cosx)/(x^(2)tan^(2)x))`, then the value of 3L is equal to

Answer

To solve the limit \( L = \lim_{x \to 0} \frac{e^{-x^2/2} - \cos x}{x^2 \tan^2 x} \), we will follow these steps:

### Step 1: Analyze the limit
As \( x \to 0 \), both the numerator and denominator approach 0, leading to a \( \frac{0}{0} \) indeterminate form. Therefore, we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Differentiate the numerator and the denominator with respect to \( x \):

- **Numerator**:
\[ \frac{d}{dx}(e^{-x^2/2} - \cos x) = -x e^{-x^2/2} + \sin x \]

- **Denominator**:
\[ \frac{d}{dx}(x^2 \tan^2 x) = 2x \tan^2 x + x^2 \cdot 2 \tan x \sec^2 x \]

Thus, we have:
\[ L = \lim_{x \to 0} \frac{-x e^{-x^2/2} + \sin x}{2x \tan^2 x + x^2 \cdot 2 \tan x \sec^2 x} \]

### Step 3: Evaluate the limit again
Substituting \( x = 0 \) directly into the new limit still gives \( \frac{0}{0} \). We apply L'Hôpital's Rule again.

### Step 4: Differentiate again
Differentiate the new numerator and denominator:

- **Numerator**:
\[ \frac{d}{dx}(-x e^{-x^2/2} + \sin x) = -e^{-x^2/2} + x^2 e^{-x^2/2} + \cos x \]

- **Denominator**:
Using the product rule and chain rule:
\[ \frac{d}{dx}(2x \tan^2 x + x^2 \cdot 2 \tan x \sec^2 x) \]
This will be more complicated, but we can simplify it later.

### Step 5: Simplify the limit
Instead of differentiating again, we can use Taylor series expansions for small \( x \):

- **For \( e^{-x^2/2} \)**:
\[ e^{-x^2/2} \approx 1 - \frac{x^2}{2} + O(x^4) \]

- **For \( \cos x \)**:
\[ \cos x \approx 1 - \frac{x^2}{2} + O(x^4) \]

Thus, the numerator becomes:
\[ e^{-x^2/2} - \cos x \approx \left(1 - \frac{x^2}{2}\right) - \left(1 - \frac{x^2}{2}\right) = O(x^4) \]

- **For \( \tan x \)**:
\[ \tan x \approx x + \frac{x^3}{3} + O(x^5) \]
So, \( \tan^2 x \approx x^2 + \frac{2x^4}{3} + O(x^6) \).

### Step 6: Substitute back into the limit
Now, substituting these approximations into the limit:
\[ L = \lim_{x \to 0} \frac{O(x^4)}{x^2 \left(x^2 + \frac{2x^4}{3}\right)} = \lim_{x \to 0} \frac{O(x^4)}{x^4 + \frac{2x^6}{3}} = \lim_{x \to 0} \frac{O(x^4)}{x^4} = \text{constant} \]

### Step 7: Final calculation
After simplifications, we find:
\[ L = \frac{1}{12} \]

### Step 8: Calculate \( 3L \)
Finally, we compute:
\[ 3L = 3 \times \frac{1}{12} = \frac{1}{4} \]

Thus, the value of \( 3L \) is \( \frac{1}{4} \).