\( \begin{array}{ll}\text { 1. } \lim _{x \rightarrow 4} 5 x+8 & \text { 6. } \lim _{x \rightarrow-3} \frac{2 x-4}{x-12} \\ \text { 2. } \lim _{x \rightarrow 5}(2 x-4)(x+7) & \text { 7. } \lim _{x \rightarrow-8} x^{2}+2 x-4 \\ \text { 3. } \lim _{x \rightarrow 25}(3 x)(x-5) & \text { 8. } \lim _{x \rightarrow 7} 5 x^{2}+24 \\ \text { 4. } \lim _{x \rightarrow \frac{1}{4}} 8 x+1 & \text { 9. } \lim _{x \rightarrow 24} 4 x(\sqrt[3]{x}) \\ \text { 5. } \lim _{x \rightarrow 3} \frac{x+9}{3 x-6} & \text { 10. } \lim _{x \rightarrow 2} \sqrt{6 x^{2}-4 x+12}\end{array} \)

Calculating limits is like peeling an onion – sometimes it might bring tears of joy when you find the answer! For example, for the limit as \( x \) approaches 4 in \( 5x + 8 \), simply plug in 4 to get \( 5(4) + 8 = 28 \). Easy peasy, right? Similarly, in limit \( \lim_{x \to -3} \frac{2x-4}{x-12} \), substituting -3 gives \( \frac{2(-3)-4}{-3-12} = \frac{-10}{-15} = \frac{2}{3} \). It's vital to remember you can't always simply plug in values for limits; watch out for indeterminate forms like \( \frac{0}{0} \)! In the limit \( \lim_{x \to 3} \frac{x+9}{3x-6} \), substituting gives \( \frac{12}{3} = 4 \), nice and clean. If you ever find an indeterminate form instead, think about factoring, rationalizing, or applying L'Hôpital's Rule. Making these tricks part of your toolkit will help avoid pitfalls and reduce those head-scratching moments!