Let's solve each limit problem provided in the image.

Problem (a)

$\lim_{{x \to 4}} \frac{{2x - 8}}{{\sqrt{x} - 2}}$

Solution:

1. Substitute $x = 4$: $\frac{{2(4) - 8}}{{\sqrt{4} - 2}} = \frac{{8 - 8}}{{2 - 2}} = \frac{0}{0}$ This is an indeterminate form, so we need to simplify further.

2. Rewrite $2x - 8$ as $2(x - 4)$: $\lim_{{x \to 4}} \frac{{2(x - 4)}}{{\sqrt{x} - 2}}$

3. Now, use the conjugate to simplify the denominator. Multiply the numerator and the denominator by $\sqrt{x} + 2$: $\lim_{{x \to 4}} \frac{{2(x - 4)(\sqrt{x} + 2)}}{{(\sqrt{x} - 2)(\sqrt{x} + 2)}}$

4. The denominator becomes a difference of squares: $\lim_{{x \to 4}} \frac{{2(x - 4)(\sqrt{x} + 2)}}{{x - 4}}$

5. Now, cancel $x - 4$ from the numerator and the denominator: $\lim_{{x \to 4}} 2(\sqrt{x} + 2)$

6. Substitute $x = 4$: $2(\sqrt{4} + 2) = 2(2 + 2) = 2 \times 4 = 8$

So, the answer for part (a) is: $\lim_{{x \to 4}} \frac{{2x - 8}}{{\sqrt{x} - 2}} = 8$

Problem (b)

$\lim_{{x \to 2}} \frac{{x^2 + 2x - 8}}{{x - 2}}$

Solution:

1. Substitute $x = 2$: $\frac{{2^2 + 2(2) - 8}}{{2 - 2}} = \frac{{4 + 4 - 8}}{{0}} = \frac{0}{0}$ This is an indeterminate form, so we need to simplify further.

2. Factor the numerator $x^2 + 2x - 8$: $x^2 + 2x - 8 = (x - 2)(x + 4)$

3. Rewrite the limit: $\lim_{{x \to 2}} \frac{{(x - 2)(x + 4)}}{{x - 2}}$

4. Cancel $x - 2$ from the numerator and the denominator: $\lim_{{x \to 2}} (x + 4)$

5. Substitute $x = 2$: $2 + 4 = 6$

So, the answer for part (b) is: $\lim_{{x \to 2}} \frac{{x^2 + 2x - 8}}{{x - 2}} = 6$

Summary

• (a) $\lim_{{x \to 4}} \frac{{2x - 8}}{{\sqrt{x} - 2}} = 8$
• (b) $\lim_{{x \to 2}} \frac{{x^2 + 2x - 8}}{{x - 2}} = 6$

Would you like further details or have any questions?

Related Questions

1. How do we handle indeterminate forms in limits?
2. What are some common methods for simplifying limits?
3. When should we use conjugates in limit problems?
4. Can you provide examples of limits involving radicals?
5. How can factoring help in solving limit problems?

Tip

When solving limits that result in $\frac{0}{0}$, consider factoring, using conjugates, or applying L'Hôpital's Rule if applicable.