Question 1 Let f(x)=4x−1f(x) = 4x - 1 Use the ε\varepsilon-δ\delta definition of the limit to prove that: limx→2f(x)=7\lim_{x \to 2} f(x) = 7 Show solutionHide solution+Question 1 - Solution We are given: f(x)=4x−1,limx→2f(x)=7f(x) = 4x - 1, \quad \lim_{x \to 2} f(x) = 7 We will prove this using the ε\varepsilon-δ\delta definition: Goal: For every ε>0\varepsilon > 0, we must find δ>0\delta > 0 such that: 0<|x−2|<δ⇒|f(x)−7|<ε0 < |x - 2| < \delta \quad \Rightarrow \quad |f(x) - 7| < \varepsilon Step 1: Write out |f(x)−7||f(x) - 7|: |f(x)−7|=|4x−1−7|=|4x−8|=4|x−2||f(x) - 7| = |4x - 1 - 7| = |4x - 8| = 4|x - 2| Step 2: Make 4|x−2|<ε4|x - 2| < \varepsilon: We can achieve this by choosing: |x−2|<ε4|x - 2| < \frac{\varepsilon}{4} Step 3: Let δ=ε4\delta = \frac{\varepsilon}{4}. Then, for all xx such that 0<|x−2|<δ0 < |x - 2| < \delta, we have: |f(x)−7|=4|x−2|<4δ=4⋅ε4=ε|f(x) - 7| = 4|x - 2| < 4\delta = 4 \cdot \frac{\varepsilon}{4} = \varepsilon Conclusion: Therefore, limx→2f(x)=7 by the ε-δ definition.\text{Therefore, } \lim_{x \to 2} f(x) = 7 \text{ by the }\varepsilon\text{-}\delta\text{ definition.}