To solve these two limit problems, let’s go through each part step-by-step.

Part (a)

$\lim_{{x \to 2}} \frac{x^2 + x}{x + 3}$

Here, we can substitute $x = 2$ directly, since there’s no issue with the denominator becoming zero:

1. Substitute $x = 2$: $\frac{2^2 + 2}{2 + 3} = \frac{4 + 2}{5} = \frac{6}{5}$

So, the solution for part (a) is: $\lim_{{x \to 2}} \frac{x^2 + x}{x + 3} = \frac{6}{5}$

Part (b)

$\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2}$

This expression is an indeterminate form ($\frac{0}{0}$) if we substitute $x = 2$. To resolve this, we need to factor the numerator.

1. Factor the numerator $x^2 - 4$ using the difference of squares: $x^2 - 4 = (x - 2)(x + 2)$

2. Substitute back into the limit: $\lim_{{x \to 2}} \frac{(x - 2)(x + 2)}{x - 2}$

3. Cancel the $x - 2$ terms: $\lim_{{x \to 2}} (x + 2)$

4. Substitute $x = 2$: $2 + 2 = 4$

So, the solution for part (b) is: $\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2} = 4$

Summary of Results

• (a) $\lim_{{x \to 2}} \frac{x^2 + x}{x + 3} = \frac{6}{5}$
• (b) $\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2} = 4$

Would you like more details on any steps or have any questions?

Related Questions:

1. What is the difference between limits that allow direct substitution and those that require factoring?
2. How does the concept of indeterminate forms apply to limit problems?
3. Can the limit of a function always be found by direct substitution?
4. How does factoring help in finding limits of rational expressions?
5. What other techniques can be used for limits when factoring is not possible?

Tip:

When encountering an indeterminate form $\frac{0}{0}$ in a limit, consider factoring, rationalizing, or using L'Hôpital's Rule if applicable.