17. \( \lim _{x \rightarrow 2} \frac{x^{2}-x-2}{x-2} \) \( \begin{array}{ll}\text { 19. } \lim _{t \rightarrow 0} \frac{t^{2}+2 t}{t^{2}-2 t} \\ \lim _{x \rightarrow-1} \frac{x^{2}+2 x+1}{x+1} & \text { 20. } \lim _{t \rightarrow 1} \frac{t^{2}-1}{t-1}\end{array} \)

To evaluate the limit \( \lim _{x \rightarrow 2} \frac{x^{2}-x-2}{x-2} \), we first factor the numerator: \( x^2 - x - 2 = (x-2)(x+1) \). Thus, our limit simplifies to \( \lim _{x \rightarrow 2} (x + 1) \), which equals 3 when we substitute \( x = 2 \). Now for \( \lim _{t \rightarrow 0} \frac{t^{2}+2t}{t^{2}-2t} \), we can factor both the numerator and the denominator: \( t(t + 2) \) and \( t(t - 2) \) respectively. Cancelling \( t \) gives us \( \lim _{t \rightarrow 0} \frac{t + 2}{t - 2} \), which evaluates to -1 when \( t = 0 \). Fun fact: Both limits involve polynomial expressions that are continuous everywhere, allowing them to be simplified easily. This illustrates the beauty of limits; they can often be resolved through clever factoring or canceling terms, making them a smooth ride rather than a bumpy road! For those wanting to dive deeper, consider exploring L'Hôpital's Rule, a powerful tool in calculus for evaluating indeterminate forms like \( \frac{0}{0} \). It's a handy method when limits lead to non-defined situations, allowing for derivatives to come to the rescue!