The value of `lim_(x to 0) (e^(nx)-(1+nx+n^2/2x^2))/x^3 (n gt 0)` is

Options

A. `n^2/6`
B. `n^3/3`
C. `n^3/6`
D. `1//6`

Answer

To solve the limit \( \lim_{x \to 0} \frac{e^{nx} - (1 + nx + \frac{n^2}{2}x^2)}{x^3} \) where \( n > 0 \), we will use the Taylor series expansion for \( e^{nx} \).

### Step-by-step Solution:

1. **Recall the Taylor Series Expansion**:
The Taylor series expansion of \( e^{nx} \) around \( x = 0 \) is:
\[ e^{nx} = 1 + nx + \frac{(nx)^2}{2!} + \frac{(nx)^3}{3!} + \cdots \]
This can be simplified to:
\[ e^{nx} = 1 + nx + \frac{n^2 x^2}{2} + \frac{n^3 x^3}{6} + O(x^4) \]

2. **Substituting the Expansion into the Limit**:
We substitute the series expansion into the limit:
\[ e^{nx} - (1 + nx + \frac{n^2}{2}x^2) = \left(1 + nx + \frac{n^2 x^2}{2} + \frac{n^3 x^3}{6} + O(x^4)\right) - \left(1 + nx + \frac{n^2}{2}x^2\right) \]
Simplifying this gives:
\[ e^{nx} - (1 + nx + \frac{n^2}{2}x^2) = \frac{n^3 x^3}{6} + O(x^4) \]

3. **Setting Up the Limit**:
Now we can rewrite the limit:
\[ \lim_{x \to 0} \frac{\frac{n^3 x^3}{6} + O(x^4)}{x^3} \]
This simplifies to:
\[ \lim_{x \to 0} \left(\frac{n^3}{6} + \frac{O(x^4)}{x^3}\right) \]

4. **Evaluating the Limit**:
As \( x \to 0 \), the term \( \frac{O(x^4)}{x^3} \) approaches \( 0 \). Therefore, we have:
\[ \lim_{x \to 0} \left(\frac{n^3}{6} + 0\right) = \frac{n^3}{6} \]

5. **Final Result**:
Thus, the value of the limit is:
\[ \frac{n^3}{6} \]

### Conclusion:
The final answer is:
\[ \lim_{x \to 0} \frac{e^{nx} - (1 + nx + \frac{n^2}{2}x^2)}{x^3} = \frac{n^3}{6} \]