Let's start solving the limits one by one from the image you uploaded:

Problem 207

$\lim_{x \to 1} \left( x^2 + 2x - 3 \right)$

To solve this, simply substitute $x = 1$ into the expression, as it is a polynomial:

$x^2 + 2x - 3 = 1^2 + 2(1) - 3 = 1 + 2 - 3 = 0$

Thus, the solution to problem 207 is: $\lim_{x \to 1} (x^2 + 2x - 3) = 0$

Problem 208

$\lim_{x \to -2} \frac{4x^2 - 3x}{x + 2}$

When substituting $x = -2$ directly into the denominator, you get zero in the denominator, which makes direct substitution invalid. Therefore, we need to simplify the expression by factoring or using another method.

The numerator is $4x^2 - 3x$. Let's factor out $x$:

$4x^2 - 3x = x(4x - 3)$

Now, the limit becomes:

$\lim_{x \to -2} \frac{x(4x - 3)}{x + 2}$

Notice that the expression is not immediately factorable to cancel out $x + 2$, and hence we need to apply limits through other techniques like L'Hopital's Rule or verify whether this leads to an infinite limit. Substituting $x = -2$ in the numerator gives:

$4(-2)^2 - 3(-2) = 4(4) + 6 = 16 + 6 = 22$

Thus, the limit is: $\lim_{x \to -2} \frac{22}{0}$

This suggests the limit tends to infinity (or negative infinity depending on the direction).

Problem 209

$\lim_{x \to 1} \frac{x^2 + \sqrt{x + 3}}{\sqrt{x + 8}}$

Substitute $x = 1$ directly:

$\frac{1^2 + \sqrt{1 + 3}}{\sqrt{1 + 8}} = \frac{1 + \sqrt{4}}{\sqrt{9}} = \frac{1 + 2}{3} = \frac{3}{3} = 1$

Thus, the solution to problem 209 is: $\lim_{x \to 1} \frac{x^2 + \sqrt{x + 3}}{\sqrt{x + 8}} = 1$

Would you like further explanations or details on any of these steps?

Here are 5 relative questions to expand your understanding:

1. How do we apply L'Hopital's Rule when direct substitution gives an indeterminate form like $\frac{0}{0}$?
2. What techniques can be used to simplify rational expressions with quadratic terms?
3. Why do limits approach infinity when the denominator tends to zero and the numerator is nonzero?
4. How does the square root function affect limit evaluation?
5. How do we evaluate limits involving absolute values?

Tip: Always check for factors that might cancel out when solving limits, especially when substitution gives an indeterminate form like $\frac{0}{0}$.