Thanks for the clarification! Let's break down and solve the corrected problems shown in your image:

1. $\lim_{x \to 0} \frac{\sin^2(x)}{x}$

Solution:

We need to carefully handle this using limit techniques or approximations.

Step 1: Use the approximation:

$\sin(x) \approx x \quad \text{for } x \approx 0$ Thus, $\sin^2(x) \approx x^2 \quad \text{for small } x.$

Step 2: Substitute the approximation:

$\lim_{x \to 0} \frac{\sin^2(x)}{x} = \lim_{x \to 0} \frac{x^2}{x} = \lim_{x \to 0} x = 0.$

2. $\lim_{x \to 0} \left( \frac{e^t - 1}{t} + \frac{x^2 + 13x + 4}{x + 3} \right)$

Let’s solve this expression term by term.

First Term: $\lim_{x \to 0} \frac{e^t - 1}{t}$

Since the variable is approaching zero and the base $e$ function is continuous and differentiable, the limit evaluates to: $\lim_{t \to 0} \frac{e^t - 1}{t} = 1.$

Second Term: $\lim_{x \to 0} \frac{x^2 + 13x + 4}{x + 3}$

Perform polynomial long division or factorization.

Step 1: Factor $x^2 + 13x + 4$: $x^2 + 13x + 4 = (x + 1)(x + 4).$

Step 2: Rewrite the expression: $\frac{(x + 1)(x + 4)}{x + 3}.$

Step 3: Evaluate the limit by direct substitution: $\lim_{x \to 0} \frac{(x + 1)(x + 4)}{x + 3} = \frac{(0 + 1)(0 + 4)}{0 + 3} = \frac{1 \times 4}{3} = \frac{4}{3}.$

Final Answer for Problem 2:

$\lim_{x \to 0} \left( \frac{e^t - 1}{t} + \frac{x^2 + 13x + 4}{x + 3} \right) = 1 + \frac{4}{3} = \frac{7}{3}.$

3. $\lim_{x \to 3} \frac{x^2 - 9}{\sqrt{x} + 3}$

Solution:

Step 1: Factor the numerator $x^2 - 9$ as a difference of squares: $x^2 - 9 = (x - 3)(x + 3).$

Step 2: Simplify the expression: $\frac{(x - 3)(x + 3)}{\sqrt{x} + 3}.$

Step 3: Cancel out the common term $x - 3$ (since $x \to 3$ avoids division by zero): $\lim_{x \to 3} \frac{(x - 3)(x + 3)}{\sqrt{x} + 3} = \lim_{x \to 3} \frac{x + 3}{\sqrt{x} + 3}.$

Step 4: Substitute $x = 3$: $\frac{3 + 3}{\sqrt{3} + 3} = \frac{6}{\sqrt{3} + 3}.$

Step 5: Rationalize the denominator: Multiply by $\frac{\sqrt{3} - 3}{\sqrt{3} - 3}$: $\frac{6(\sqrt{3} - 3)}{(\sqrt{3} + 3)(\sqrt{3} - 3)} = \frac{6(\sqrt{3} - 3)}{3 - 9} = \frac{6(\sqrt{3} - 3)}{-6} = -(\sqrt{3} - 3) = 3 - \sqrt{3}.$

Final Answers:

1. $\lim_{x \to 0} \frac{\sin^2(x)}{x} = 0$
2. $\lim_{x \to 0} \left( \frac{e^t - 1}{t} + \frac{x^2 + 13x + 4}{x + 3} \right) = \frac{7}{3}$
3. $\lim_{x \to 3} \frac{x^2 - 9}{\sqrt{x} + 3} = 3 - \sqrt{3}$

Would you like more steps or a deeper explanation?

Related Questions:

1. How can L'Hôpital's rule be applied to these problems?
2. What are common techniques for solving indeterminate forms in limits?
3. How does factorization help in solving rational function limits?
4. What is the significance of the exponential function in limit evaluations?
5. How can rationalizing the denominator simplify expressions?

Tip:

Always check for indeterminate forms like $\frac{0}{0}$ or (\in