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

1. The Maclaurin Series for $e^{x}$

The Maclaurin series expansion of $e^x$ is:
$e^x = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$

Now, replace $x$ with $x^2$ to expand $e^{x^2}$:
$e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \dots$

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

From the series for $e^{x^2}$, subtract $x^2$ and $1$:
$e^{x^2} - x^2 - 1 = \left( 1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6} + \dots \right) - x^2 - 1.$

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

3. Divide by $4x^4$:

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

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

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

4. Take the Limit as $x \to 0$:

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

Final Answer:

$\boxed{\frac{1}{8}}$

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

Related Questions:

1. How do you derive the Maclaurin series for $e^x$?
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.