\( \begin{array}{ll}\text { 1. }-\lim _{x \rightarrow 2} x^{4} & \text { 8. }-\lim _{x \rightarrow 4} \frac{4-x}{x^{2}-16} \\ \text { 2.- } \lim _{x \rightarrow 0}(2 x-1) & \text { 9. }-\lim _{x \rightarrow 2} \frac{x^{4}-16}{x-2} \\ \text { 3. }-\lim _{x \rightarrow-3}\left(x^{2}+3 x\right) & \text { 10. } \lim _{x \rightarrow 5} \frac{x^{2}-25}{x-5} \\ \text { 4. }-\lim _{x \rightarrow-3}\left(2 x^{2}+4 x+1\right) & \text { 11. }-\lim _{x \rightarrow-4} \frac{16-x^{2}}{4+x} \\ \text { 5. }-\lim _{x \rightarrow 2} \frac{1}{x} & \text { 12. }-\lim _{x \rightarrow 1} \frac{x^{2}+2 x-3}{x-1} \\ \text { 6. }-\lim _{x \rightarrow 1} \frac{x-3}{x^{2}+4} & \text { 13. }-\lim _{x \rightarrow 1} \frac{1-x^{2}}{x^{2}+x-2} \\ \text { 7. }-\lim _{x \rightarrow 7} \frac{5 x}{\sqrt{x+2}} & \end{array} \)

Understanding limits is like having a magic key that opens up the world of calculus! It allows you to predict the behavior of functions as they approach specific locations. For example, when looking at a function like \(f(x) = \frac{x^2 - 25}{x - 5}\), evaluating the limit as \(x\) approaches 5 reveals interesting behavior, even though direct substitution results in an undefined form. This is often where rewriting the function or factoring comes into play! When you confront limits that result in indeterminate forms, one common mistake is to ignore simplification steps. For instance, in the problem involving \( \lim_{x \rightarrow 4} \frac{4 - x}{x^2 - 16} \), skipping factorization can lead to confusion. Always remember, don't just plug in; use algebraic manipulation to simplify and check for cancellations. This way, you'll avoid those pesky errors and get to the smoother solutions much faster!