lim x tends to zero [e^x - ( 1 + x + (x^2/2))]

We are tasked with evaluating the limit:

\[
\lim_{x \to 0} \left( e^x - \left(1 + x + \frac{x^2}{2} \right) \right)
\]

### Step-by-Step Solution:

1. **Recall the Taylor expansion of \( e^x \) around \( x = 0 \):**

   \[
   e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots
   \]

2. **Subtract the polynomial \( 1 + x + \frac{x^2}{2} \) from \( e^x \):**

   \[
   e^x - \left( 1 + x + \frac{x^2}{2} \right) = \left( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \right) - \left( 1 + x + \frac{x^2}{2} \right)
   \]

   Simplifying this expression:

   \[
   e^x - \left( 1 + x + \frac{x^2}{2} \right) = \frac{x^3}{6} + \frac{x^4}{24} + \cdots
   \]

3. **Take the limit as \( x \to 0 \):**

   Since each term in the series starting from \( \frac{x^3}{6} \) involves powers of \( x \) higher than 2, all these terms vanish as \( x \to 0 \).

   \[
   \lim_{x \to 0} \left( \frac{x^3}{6} + \frac{x^4}{24} + \cdots \right) = 0
   \]

Thus, the limit is:

\[
\boxed{0}
\]

Let me know if you need further details or clarifications! 

### Related questions:
1. What is the Taylor series of \( \sin(x) \) around \( x = 0 \)?
2. How can we apply L'Hopital's rule to limits involving indeterminate forms?
3. What is the general form of a Taylor series?
4. How do we expand \( \ln(1+x) \) around \( x = 0 \)?
5. Can you provide an example of a limit using a different function, such as \( \cos(x) \)?

**Tip**: Taylor expansions are useful for approximating functions near a specific point, especially when dealing with limits!

Explore the limit of e^x - (1 + x + (x^2/2)) as x tends to 0 using Taylor expansion. This detailed solution breaks down the key steps and concepts in limits and series expansion. Ideal for students in Grades 11-12 or introductory calculus courses.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation