Thank you for providing the image and the clarification. Here's the updated problem statement based on your instructions:

Updated Problem 2:

$\lim_{x \to 0} \frac{e^t - 1}{t} + \frac{x^2 + 3x + 4}{x + 3}$

I'll now solve the updated second problem step by step.

Solution for Updated Problem 2:

1. First Term: $\frac{e^t - 1}{t}$ as $t \to 0$:

   • Using the Taylor expansion of $e^t$ around $t = 0$: $e^t \approx 1 + t + \frac{t^2}{2} + \cdots$ Substituting this into $\frac{e^t - 1}{t}$: $\frac{e^t - 1}{t} = \frac{(1 + t + \frac{t^2}{2} + \cdots) - 1}{t} = \frac{t + \frac{t^2}{2} + \cdots}{t} = 1 + \frac{t}{2} + \cdots$ As $t \to 0$, the higher-order terms vanish, leaving: $\lim_{t \to 0} \frac{e^t - 1}{t} = 1.$

2. Second Term: $\frac{x^2 + 3x + 4}{x + 3}$ as $x \to 0$:

   • Factorize the numerator if possible: $x^2 + 3x + 4 \text{ cannot be factored easily}.$
   • Direct substitution: $\frac{x^2 + 3x + 4}{x + 3} = \frac{0^2 + 3(0) + 4}{0 + 3} = \frac{4}{3}.$

3. Combine Both Results:

   $$$$

From the two terms calculated:

1. $\lim_{t \to 0} \frac{e^t - 1}{t} = 1$
2. $\lim_{x \to 0} \frac{x^2 + 3x + 4}{x + 3} = \frac{4}{3}$

Now combine the results: [ \text{Total Limit} = 1 + \frac{4}{3} = \frac{3}{3} + \frac{4}{3} = \frac{7}{3}.$$

Final Answer for Updated Problem 2:

$\lim_{x \to 0} \frac{e^t - 1}{t} + \frac{x^2 + 3x + 4}{x + 3} = \frac{7}{3}.$

Would you like me to solve the other problems or explain anything in more detail?

Related Questions:

1. How is the Taylor expansion of $e^t$ derived?
2. What are the conditions for direct substitution in limit problems?
3. Can $\frac{x^2 + 3x + 4}{x + 3}$ be simplified further using long division?
4. How do higher-order terms affect the limit as $t \to 0$?
5. What happens if $t$ and $x$ approach different values (not zero)?

Tip:

When dealing with limits, check for opportunities to use Taylor expansions or direct substitution before attempting more complex methods.