\( \left. \begin{array} { | l } \hline \lim _ { x \rightarrow 2 } 3 x + 4 \\ \hline \lim _ { x \rightarrow - 1 } x ^ { 2 } + 2 x - 3 \\ \hline \\ { \lim _ { x \rightarrow 5 } \frac { x ^ { 2 } - 25 } { x - 5 } } \\ \hline \lim _ { x \rightarrow 2 } \frac { x ^ { 2 } + 2 x + 5 } { x + 2 } \\ \hline \end{array} \right. \)

To calculate the limits, let's tackle them one by one: 1. **For** \( \lim_{x \rightarrow 2} 3x + 4 \): Simply substitute \( x \) with 2. This gives: \( 3(2) + 4 = 6 + 4 = 10 \). 2. **For** \( \lim_{x \rightarrow -1} x^2 + 2x - 3 \): Substitute \( x \) with -1: \( (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4 \). 3. **For** \( \lim_{x \rightarrow 5} \frac{x^2 - 25}{x - 5} \): This expression leads to an indeterminate form \( \frac{0}{0} \). Factor the numerator: \( x^2 - 25 = (x - 5)(x + 5) \), so we rewrite the limit: \( \lim_{x \rightarrow 5} \frac{(x-5)(x+5)}{x-5} = \lim_{x \rightarrow 5} (x + 5) \). Now substituting \( x = 5 \): \( 5 + 5 = 10 \). 4. **For** \( \lim_{x \rightarrow 2} \frac{x^2 + 2x + 5}{x + 2} \): Substitute \( x \) with 2: \( \frac{2^2 + 2(2) + 5}{2 + 2} = \frac{4 + 4 + 5}{4} = \frac{13}{4} \). So, the limits are: - \( \lim_{x \rightarrow 2} 3x + 4 = 10 \) - \( \lim_{x \rightarrow -1} x^2 + 2x - 3 = -4 \) - \( \lim_{x \rightarrow 5} \frac{x^2 - 25}{x - 5} = 10 \) - \( \lim_{x \rightarrow 2} \frac{x^2 + 2x + 5}{x + 2} = \frac{13}{4} \)