The image contains three limit problems. Let’s go through each one in turn:

(b) $\lim_{{x \to 0}} \frac{x^2 - 7x}{x}$

1. First, simplify the expression by factoring the numerator: $x^2 - 7x = x(x - 7)$ So the expression becomes: $\frac{x(x - 7)}{x}$
2. Cancel the $x$ terms (assuming $x \neq 0$): $x - 7$
3. Now, take the limit as $x \to 0$: $\lim_{{x \to 0}} (x - 7) = 0 - 7 = -7$

Thus, the result for part (b) is: $\lim_{{x \to 0}} \frac{x^2 - 7x}{x} = -7$

(c) $\lim_{{x \to 0^+}} \frac{x^2 - 4x}{x}$

1. Simplify the expression similarly by factoring the numerator: $x^2 - 4x = x(x - 4)$ So, the expression becomes: $\frac{x(x - 4)}{x}$
2. Cancel the $x$ terms (assuming $x \neq 0$): $x - 4$
3. Now, take the limit as $x \to 0^+$: $\lim_{{x \to 0^+}} (x - 4) = 0 - 4 = -4$

Thus, the result for part (c) is: $\lim_{{x \to 0^+}} \frac{x^2 - 4x}{x} = -4$

Let me know if you want any further details or if anything needs clarification!

Related Questions:

1. What happens if the limits are approached from both sides instead of just the right side?
2. How does factoring help simplify the process of calculating limits?
3. Are there any conditions where factoring wouldn’t work for simplifying limits?
4. Can we use L'Hôpital's rule to solve these problems?
5. How do these problems change if the powers of $x$ in the numerator were higher?

Tip:

When evaluating limits, always check if direct substitution leads to an indeterminate form (like $\frac{0}{0}$), and use techniques like factoring or L'Hôpital's Rule accordingly.