Q5 - Use the precise definition of the limit to prove the following limit. \[ \lim _{x \rightarrow 2}(5 x+8)=18 \]

1. **Given statement** We want to prove that \[ \lim_{x \to 2}(5x + 8) = 18. \] This means that according to the precise definition of the limit, for every \( \epsilon > 0 \) there exists a \( \delta > 0 \) such that if \[ 0 < |x - 2| < \delta, \] then \[ |(5x + 8) - 18| < \epsilon. \] 2. **Start with the expression inside the absolute value** Compute the difference: \[ |(5x + 8) - 18| = |5x + 8 - 18| = |5x - 10|. \] 3. **Factor the expression** Factor out the common factor: \[ |5x - 10| = 5|x - 2|. \] 4. **Set the inequality condition** We want: \[ 5|x - 2| < \epsilon. \] To achieve this, we need: \[ |x - 2| < \frac{\epsilon}{5}. \] 5. **Choose \(\delta\)** Let \[ \delta = \frac{\epsilon}{5}. \] Then, whenever \[ |x - 2| < \delta, \] it follows that \[ |5x+8 - 18| = 5|x-2| < 5\left(\frac{\epsilon}{5}\right) = \epsilon. \] 6. **Conclusion** By the definition of the limit, since for every \( \epsilon > 0 \) we have found a corresponding \( \delta = \frac{\epsilon}{5} \) such that \[ |x - 2| < \delta \implies |(5x+8) - 18| < \epsilon, \] we conclude that \[ \lim_{x\to 2}(5x+8)=18. \]