The value of `lim_(xrarr0)(1-cos^(3)x)/(sin^(2)xcos x)` is equal to

Answer

To solve the limit \( \lim_{x \to 0} \frac{1 - \cos^3 x}{\sin^2 x \cos x} \), we will follow these steps:

### Step 1: Rewrite the limit
We start with the limit expression:
\[ \lim_{x \to 0} \frac{1 - \cos^3 x}{\sin^2 x \cos x} \]

### Step 2: Use the identity for \(1 - \cos^3 x\)
We can factor \(1 - \cos^3 x\) using the identity \(1 - a^3 = (1 - a)(1 + a + a^2)\):
\[ 1 - \cos^3 x = (1 - \cos x)(1 + \cos x + \cos^2 x) \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{(1 - \cos x)(1 + \cos x + \cos^2 x)}{\sin^2 x \cos x} \]

### Step 3: Substitute \(1 - \cos x\) using Taylor series
Using the Taylor series expansion, we know that:
\[ 1 - \cos x \approx \frac{x^2}{2} \quad \text{as } x \to 0 \]
Substituting this into our limit gives:
\[ \lim_{x \to 0} \frac{\frac{x^2}{2}(1 + \cos x + \cos^2 x)}{\sin^2 x \cos x} \]

### Step 4: Substitute \(\sin^2 x\) and \(\cos x\)
We also know that:
\[ \sin x \approx x \quad \text{as } x \to 0 \implies \sin^2 x \approx x^2 \]
And \(\cos x \approx 1\) as \(x \to 0\). Therefore, we can substitute:
\[ \sin^2 x \approx x^2 \quad \text{and} \quad \cos x \approx 1 \]
Now our limit becomes:
\[ \lim_{x \to 0} \frac{\frac{x^2}{2}(1 + 1 + 1)}{x^2 \cdot 1} = \lim_{x \to 0} \frac{\frac{x^2}{2} \cdot 3}{x^2} \]

### Step 5: Simplify the limit
The \(x^2\) terms cancel out:
\[ \lim_{x \to 0} \frac{3}{2} = \frac{3}{2} \]

### Conclusion
Thus, the value of the limit is:
\[ \frac{3}{2} \]