To solve these limits using power series, we can expand the functions in the numerator and denominator using their Taylor series at $x = 0$ and then evaluate the limit based on the leading terms.

Let's go through each part one by one.

Part (a) $\lim_{x \to 0} \frac{1 + x - e^x}{x^2}$

1. Expand $e^x$ in a Taylor series around $x = 0$: $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots$

2. Substitute this expansion into the expression: $1 + x - e^x = 1 + x - \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \right)$ Simplify by canceling terms: $1 + x - e^x = -\frac{x^2}{2} - \frac{x^3}{6} - \cdots$

3. Rewrite the limit: $\lim_{x \to 0} \frac{1 + x - e^x}{x^2} = \lim_{x \to 0} \frac{-\frac{x^2}{2} - \frac{x^3}{6} - \cdots}{x^2}$

4. Divide each term by $x^2$: $= \lim_{x \to 0} \left(-\frac{1}{2} - \frac{x}{6} - \cdots \right)$

5. Take the limit as $x \to 0$: $= -\frac{1}{2}$

So, the answer for part (a) is: $\lim_{x \to 0} \frac{1 + x - e^x}{x^2} = -\frac{1}{2}$

Part (b) $\lim_{x \to 0} \frac{x}{e^x - 1}$

1. Expand $e^x$ as before: $e^x = 1 + x + \frac{x^2}{2} + \cdots$ So, $e^x - 1 = x + \frac{x^2}{2} + \cdots$

2. Rewrite the limit: $\lim_{x \to 0} \frac{x}{x + \frac{x^2}{2} + \cdots}$

3. Factor $x$ out from the denominator: $= \lim_{x \to 0} \frac{x}{x \left(1 + \frac{x}{2} + \cdots \right)} = \lim_{x \to 0} \frac{1}{1 + \frac{x}{2} + \cdots}$

4. Take the limit as $x \to 0$: $= \frac{1}{1} = 1$

So, the answer for part (b) is: $\lim_{x \to 0} \frac{x}{e^x - 1} = 1$

Part (c) $\lim_{x \to 0} \frac{\ln(x+1)}{\sqrt{4 + x} - 2}$

1. Expand $\ln(x+1)$ using its Taylor series at $x = 0$: $\ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$

2. Expand $\sqrt{4 + x}$ using the binomial series around $x = 0$: $\sqrt{4 + x} = 2 + \frac{x}{4} - \frac{x^2}{64} + \cdots$ Therefore, $\sqrt{4 + x} - 2 = \frac{x}{4} - \frac{x^2}{64} + \cdots$

3. Rewrite the limit: $\lim_{x \to 0} \frac{x - \frac{x^2}{2} + \cdots}{\frac{x}{4} - \frac{x^2}{64} + \cdots}$

4. Factor $x$ from both the numerator and denominator: $= \lim_{x \to 0} \frac{x \left(1 - \frac{x}{2} + \cdots \right)}{x \left(\frac{1}{4} - \frac{x}{64} + \cdots \right)} = \lim_{x \to 0} \frac{1 - \frac{x}{2} + \cdots}{\frac{1}{4} - \frac{x}{64} + \cdots}$

5. Take the limit as $x \to 0$: $= \frac{1}{\frac{1}{4}} = 4$

So, the answer for part (c) is: $\lim_{x \to 0} \frac{\ln(x+1)}{\sqrt{4 + x} - 2} = 4$

Would you like further clarification on any of these parts?

Follow-Up Questions:

1. How can power series approximations be applied in calculus beyond limit calculations?
2. What are the conditions required for a power series to converge?
3. How does the Taylor series differ from the Maclaurin series?
4. Can power series be used to approximate values for non-polynomial functions?
5. Why is factoring useful when simplifying expressions involving power series?

Tip:

Remember, in limit problems, look for ways to factor or cancel terms to simplify calculations, especially with power series.