`lim_(x to 0) (sin x - x + 1/6x^(3))/x^(5)` = ……………

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sin x - x + \frac{1}{6}x^3}{x^5} \), we will use the Taylor series expansion for \( \sin x \).

### Step 1: Expand \( \sin x \)
The Taylor series expansion of \( \sin x \) around \( x = 0 \) is:
\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots \]
Substituting this into our limit gives:
\[ \sin x - x = -\frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \ldots \]

### Step 2: Substitute into the limit
Now, substituting this expansion into the limit expression:
\[ \sin x - x + \frac{1}{6}x^3 = -\frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \frac{1}{6}x^3 \]
The \( -\frac{x^3}{6} \) and \( \frac{1}{6}x^3 \) cancel each other out:
\[ = \frac{x^5}{120} - \frac{x^7}{5040} + \ldots \]

### Step 3: Rewrite the limit
Now, we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\frac{x^5}{120} - \frac{x^7}{5040} + \ldots}{x^5} \]
This simplifies to:
\[ \lim_{x \to 0} \left( \frac{1}{120} - \frac{x^2}{5040} + \ldots \right) \]

### Step 4: Evaluate the limit
As \( x \to 0 \), the terms involving \( x \) vanish:
\[ \lim_{x \to 0} \left( \frac{1}{120} - \frac{x^2}{5040} + \ldots \right) = \frac{1}{120} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\sin x - x + \frac{1}{6}x^3}{x^5} = \frac{1}{120} \]

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