Math Problem Statement

Use the Maclaurin series for \( f(x) = e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \) to evaluate the limit: \[ \lim_{x \to 0} \frac{e^{x^2} - x^2 - 1}{4x^4}. \] Hint: Write out the first few terms.

Solution

To evaluate the limit
limx0ex2x214x4\lim_{x \to 0} \frac{e^{x^2} - x^2 - 1}{4x^4}
using the Maclaurin series for exe^x, let’s proceed step by step.


1. The Maclaurin Series for exe^{x}

The Maclaurin series expansion of exe^x is:
ex=n=0xnn!=1+x+x22!+x33!+x44!+e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots

Now, replace xx with x2x^2 to expand ex2e^{x^2}:
ex2=1+x2+x42!+x63!+e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots


2. Simplify the Numerator ex2x21e^{x^2} - x^2 - 1:

From the series for ex2e^{x^2}, subtract x2x^2 and 11:
ex2x21=(1+x2+x42+x66+)x21.e^{x^2} - x^2 - 1 = \left( 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6} + \dots \right) - x^2 - 1.

Simplify:
ex2x21=x42+x66+e^{x^2} - x^2 - 1 = \frac{x^4}{2} + \frac{x^6}{6} + \dots


3. Divide by 4x44x^4:

We now divide the entire expression by 4x44x^4:
ex2x214x4=x42+x66+4x4.\frac{e^{x^2} - x^2 - 1}{4x^4} = \frac{\frac{x^4}{2} + \frac{x^6}{6} + \dots}{4x^4}.

Separate the terms:
ex2x214x4=x424x4+x664x4+\frac{e^{x^2} - x^2 - 1}{4x^4} = \frac{x^4}{2 \cdot 4x^4} + \frac{x^6}{6 \cdot 4x^4} + \dots

Simplify each term:
ex2x214x4=18+x224+\frac{e^{x^2} - x^2 - 1}{4x^4} = \frac{1}{8} + \frac{x^2}{24} + \dots


4. Take the Limit as x0x \to 0:

As x0x \to 0, the higher-order terms involving x2x^2 vanish:
limx0ex2x214x4=18.\lim_{x \to 0} \frac{e^{x^2} - x^2 - 1}{4x^4} = \frac{1}{8}.


Final Answer:

18\boxed{\frac{1}{8}}

Would you like a detailed explanation of any specific step? Let me know!


Related Questions:

  1. How do you derive the Maclaurin series for exe^x?
  2. What is the general form of the Maclaurin series for any function?
  3. How can the limit of indeterminate forms involving series be solved?
  4. What are other applications of the Maclaurin series?
  5. How can higher-order terms affect the convergence of series approximations?

Tip: Always expand the function carefully to include sufficient terms to evaluate limits accurately.

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Math Problem Analysis

Mathematical Concepts

Limits
Maclaurin Series
Calculus

Formulas

Maclaurin series for e^x: \( e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \)
Limit simplification using series expansion

Theorems

Maclaurin Series Expansion
Limit Theorem

Suitable Grade Level

College Level (Calculus I/II)