Question

Consider the line \(f(x)=m x+b,\) where \(m \neq 0 .\) Use the \(\varepsilon-\delta\) definition of a limit to prove that \(\lim f(x)=m c+b\).

Short answer

The proposed limit \(\lim_{x \to c} f(x) = mc + b\) for the line equation \(f(x) = mx + b\) where \(m \neq 0\), can be proven using the epsilon-delta definition of the limit.

Step-by-step solution

Identify the Limit

First, identify what the limit is in the constituted formula. Given that \(f(x) = mx + b\), when \(x\) goes to any real number \(c\), \(f(c) = mc + b\). Therefore, \(\lim_{x \to c} f(x) = mc + b\).

Apply the Epsilon-Delta Limit Definition

Next, resort to the epsilon-delta definition of a limit. Accordingly, for every epsilon \(> 0\), there exists a delta \(> 0\) such that if \(0 <|x-c| < \delta\) then \(|f(x) - L| < \epsilon\). Here, \(L\) is the limit that is \(mc + b\) in this case.

Substitute the Function

After the above, substitute \(f(x)\) with \(mx + b\) to obtain \(|mx + b - mc - b| < \epsilon\). This simplifies to \(|m(x - c)| < \epsilon\).

Simplify further and Choose Delta

Given that \(m ≠ 0\), it's feasible to simplify the inequality to \(|x - c| < \epsilon/|m|\). To guarantee this inequality is valid, let's choose our delta such that \(\delta = \epsilon/|m|\). This assures for \(0 <|x-c| < \delta\), it will result in \(|mx + b - mc - b| < \epsilon\), thereby proving the limit.