In formally proving that $\lim _{x \rightarrow-6}\left(\frac{1}{5} x+4\right)=\frac{14}{5}$, let $\varepsilon> 0$ be arbitrary. Determine $\delta$ as a function of $\varepsilon$.
Note: in this case $\delta$ will be a function of $\varepsilon$. You will need to write the word epsilon for $\varepsilon$ in the answerbox
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Step 1 ：Given the function \(f(x) = \frac{1}{5}x + 4\), we want to prove that \(\lim _{x \rightarrow-6}\left(\frac{1}{5} x+4\right)=\frac{14}{5}\).

Step 2 ：According to the limit definition, for every \(\varepsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x - a| < \delta\), then \(|f(x) - L| < \varepsilon\). In this case, \(a = -6\) and \(L = \frac{14}{5}\).

Step 3 ：We start by setting up the inequality \(|f(x) - L| < \varepsilon\) and solve for \(|x - a|\) to find \(\delta\).

Step 4 ：We have \(|f(x) - L| = |(1/5)x + 4 - 14/5| = |(1/5)x - 2/5| = |x - (-6)|/5\). We want this to be less than epsilon, so \(|x - (-6)|/5 < \varepsilon\), which gives \(|x - (-6)| < 5\varepsilon\).

Step 5 ：Therefore, \(\delta\) should be \(5\varepsilon\).

Step 6 ：\(\boxed{\delta = 5\varepsilon}\)