If `L=lim_(xto0) (e^(-x^(2)//2)-cosx)/(x^(3)sinx)`, then the value of `1//(3L)` is ________.

Explanation

To solve the limit \( L = \lim_{x \to 0} \frac{e^{-\frac{x^2}{2}} - \cos x}{x^3 \sin x} \), we will follow these steps:

### Step 1: Expand \( e^{-\frac{x^2}{2}} \) and \( \cos x \) using Taylor series
The Taylor series expansion of \( e^{-\frac{x^2}{2}} \) around \( x = 0 \) is:
\[ e^{-\frac{x^2}{2}} = 1 - \frac{x^2}{2} + \frac{x^4}{4!} + O(x^6) \]
The Taylor series expansion of \( \cos x \) around \( x = 0 \) is:
\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{4!} + O(x^6) \]

### Step 2: Substitute the expansions into the limit expression
Substituting these expansions into the limit gives:
\[ e^{-\frac{x^2}{2}} - \cos x = \left(1 - \frac{x^2}{2} + \frac{x^4}{4!} + O(x^6)\right) - \left(1 - \frac{x^2}{2} + \frac{x^4}{4!} + O(x^6)\right) \]
The linear terms cancel out, so we need to look at the next higher order terms:
\[ e^{-\frac{x^2}{2}} - \cos x = O(x^6) \]

### Step 3: Simplify the numerator
Since both series are equal up to \( O(x^4) \), we find that:
\[ e^{-\frac{x^2}{2}} - \cos x = \frac{x^4}{24} + O(x^6) \]

### Step 4: Expand \( \sin x \) using Taylor series
The Taylor series expansion of \( \sin x \) around \( x = 0 \) is:
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]

### Step 5: Substitute into the limit expression
Now substituting into the limit:
\[ x^3 \sin x = x^3 \left(x - \frac{x^3}{6} + O(x^5)\right) = x^4 - \frac{x^6}{6} + O(x^8) \]

### Step 6: Write the limit expression
Now we can write the limit:
\[ L = \lim_{x \to 0} \frac{\frac{x^4}{24} + O(x^6)}{x^4 - \frac{x^6}{6} + O(x^8)} \]

### Step 7: Factor out \( x^4 \)
Factoring out \( x^4 \) from both the numerator and denominator:
\[ L = \lim_{x \to 0} \frac{\frac{1}{24} + O(x^2)}{1 - \frac{x^2}{6} + O(x^4)} \]

### Step 8: Evaluate the limit
As \( x \to 0 \), the higher order terms vanish:
\[ L = \frac{\frac{1}{24}}{1} = \frac{1}{24} \]

### Step 9: Calculate \( \frac{1}{3L} \)
Now, we can find \( \frac{1}{3L} \):
\[ \frac{1}{3L} = \frac{1}{3 \cdot \frac{1}{24}} = \frac{24}{3} = 8 \]

Thus, the value of \( \frac{1}{3L} \) is \( \boxed{8} \).

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